Answer:
(a) P (X ≥ 4) = 0.972
(b) E (X) = 20
Step-by-step explanation:
Let X = number of people tested to detect the presence of gene in 2.
Then the random variable X follows a Negative binomial distribution with parameters r (number of success) and p probability of success.
The probability distribution function of X is:
[tex]f(x)={x-1\choose r-1}p^{r}(1-p)^{x-r}[/tex]
Given: r = 2 and p = 0.1
(a)
Compute the probability that four or more people will have to be tested before two with the gene are detected as follows:
P (X ≥ 4) = 1 - P (X = 3) - P (X = 2)
[tex]=1-[{3-1\choose 2-1}(0.1)^{2}(1-0.1)^{3-2}]-[{2-1\choose 2-1}(0.1)^{2}(1-0.1)^{2-2}]\\=1-0.018-0.01\\=0.972[/tex]
Thus, the probability that four or more people will have to be tested before two with the gene are detected is 0.972.
(b)
The expected value of a negative binomial random variable X is:
[tex]E(X)=\frac{r}{p}[/tex]
The expected number of people to be tested before two with gene are detected is:
[tex]E(X)=\frac{r}{p}=\frac{2}{0.1}=20[/tex]
Thus, the expected number of people to be tested before two with gene are detected is 20.
What odds should a person give in favor of the following events? (a) A card chosen at random from a 52-card deck is an ace. (b) Two heads will turn up when a coin is tossed twice. (c) Boxcars (two sixes) will turn up when two dice are rolled
Answer:
(a) 7.69%
(b) 25%
(c) 2.78%
Step-by-step explanation:
(a)
In a deck of 52 cards there are 4 aces.
The odds in favor or the probability of selecting an ace is:
[tex]P(Ace) = \frac{Number\ of\ aces}{Number\ of\ cards\ in\ total}\\ =\frac{4}{52}\\ =0.076923\\\approx7.69\%[/tex]
Thus, the probability of selecting an ace from a random deck of 562 cards is 7.69%.
(b)
The outcomes of each toss of a coin is independent of the other, since the result of the previous toss does not affect the result of the current toss.
The probability that both the tosses will end up in heads is:
[tex]P(2\ Heads)=P(1^{st}\ Head)\times P(2^{nd}\ Head)\\=\frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{4}\\ =0.25\ or\ 25\%\\[/tex]
Thus, the probability that both the tosses will end up in heads is 25%.
(c)
The sample space of two dice consists of 36 outcomes in total.
Out of these 36 outcomes there is only 1 Boxcar, i.e. two sixes.
The probability of a boxcar when two dice are rolled is:
[tex]P(Boxcar)=\frac{Favorable\ outcomes}{Total\ no.\ of\ outcomes}\\= \frac{1}{36}\\ =0.027777\\\approx2.78\%[/tex]
Thus, the probability of a boxcar when two dice are rolled is 2.78%.
Final answer:
To find the odds in favor of specific events in probability theory, one must compare the number of successful outcomes to the number of unsuccessful ones. For selecting an ace from a deck of cards, the odds are 1:12; for getting two heads from two coin tosses, the odds are 1:3; and for rolling two sixes with two dice, the odds are 1:35.
Explanation:
The question asks for the odds in favor of several different probabilistic events, which relate to the field of probability theory within mathematics. Here's how to calculate the odds for each of the requested scenarios:
(a) Odds in favor of a card being an ace: There are 4 aces in a standard 52-card deck. The odds in favor are the number of ways the event can occur (4 aces) to the number of ways the event can fail to occur (52 - 4 = 48 non-aces), which simplifies to 1:12.
(b) Odds in favor of two heads when a coin is tossed twice: The probability of getting a head on one coin toss is 1/2, and since the two tosses are independent, the probability of getting two heads is (1/2) * (1/2) = 1/4. The odds in favor are calculated by taking the probability of the event occurring (1 chance) against the probability of it not occurring (3 chances), which gives us odds of 1:3.
(c) Odds in favor of rolling boxcars (two sixes) with two dice: Each die has a 1/6 chance of rolling a six, so the probability of rolling two sixes is (1/6) * (1/6) = 1/36. The odds in favor are the number of successful outcomes (1) against the number of all other outcomes (35), resulting in odds of 1:35.
The time to fly between New York City and Chicago is uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes.
What is the probability that a flight is between 125 and 140 minutes?
A. 1.00.
B. 0.50.
C. 0.33.
D. 0.12.
E. 0.15
Answer:
B. 0.50.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability of a measure X being between two values c and d, in which d is larger than c, is given by the following formula:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
Uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes.
This means that [tex]a = 120, b = 150[/tex]
What is the probability that a flight is between 125 and 140 minutes?
This is
[tex]P(125 \leq X \leq 140) = \frac{140 - 125}{150 - 120} = 0.5[/tex]
So the correct answer is:
B. 0.50.
Same-sex unions increasingly become heated political issue. the 2006 GSS asked respondents opinions on homosexual relations. five response categories ranged from always wrong to not wrong at all. see the following frequency distributions. at what level is this variable measured? Homosexual Relations Frequency Percentage Cumulative Percentage Always wrong 467 50.2 50.2Almost always wrong 41 4.4 54.6 Sometimes wrong 76 8.2 62.8Not wrong at all 346 37.2 100.0 Total 930 100.0
Answer:
Ordinal level
Step-by-step explanation:
The variable of interest is opinion on homosexual relations and the frequency distribution for opinion on homosexual relations is given.
The opinion of people is categorized from wrong to not wrong at all. There exists order in the categorizes and measurement of variable indicates the ordinal level of measurement.
Thus, variable is measured at ordinal level.
According to a posting on a website subsequent to the death of a child who bit into a peanut, a certain study found that 7% of children younger than 18 in the United States have at least one food allergy. Among those with food allergies, about 41% had a history of severe reaction.a. If a child younger than 18 is randomly selected, what is the probability that he or she has at least one food allergy and a history of severe reaction? (Enter your answer to four decimal places.) b. It was also reported that 30% of those with an allergy in fact are allergic to multiple foods. If a child younger than 18 is randomly selected, what is the probability that he or she is allergic to multiple foods? (Enter your answer to three decimal places.)
a) The probability that he or she has at least one food allergy and a history of severe reaction is 0.0287.
b) The probability that he or she is allergic to multiple foods is, 0.021
Given that;
A certain study found that 7% of children younger than 18 in the United States have at least one food allergy.
a. Since the probability that a child younger than 18 has at least one food allergy is given as 7%.
Among those with food allergies, the probability of having a history of severe reaction is 41%.
Hence for the probability that a child has both at least one food allergy and a history of severe reaction, multiply these probabilities together:
7% × 41% = 0.07 × 0.41
= 0.0287.
Therefore, the probability is 0.0287.
b) For the probability that a randomly selected child younger than 18 is allergic to multiple foods, consider the information given.
The probability of having at least one food allergy among children younger than 18 is 7%.
And among those with allergies, 30% are allergic to multiple foods.
Hence for the probability, multiply the probability of having at least one food allergy (7%) by the probability of being allergic to multiple foods (30% of those with allergies):
Probability = 7% × 30%
= 0.07 × 0.30
= 0.021.
Therefore, the probability that a randomly selected child younger than 18 is allergic to multiple foods is 0.021.
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The probability that a child younger than 18 has at least one food allergy and a history of severe reaction is approximately 0.029. The probability that a child younger than 18 is allergic to multiple foods is approximately 0.021.
Explanation:To find the probability that a child younger than 18 has at least one food allergy and a history of severe reaction, we can use the information provided. We know that 7% of children younger than 18 have at least one food allergy and among those with food allergies, 41% had a history of severe reaction. To calculate the probability, we multiply these two probabilities together: 0.07 (the probability of having a food allergy) multiplied by 0.41 (the probability of having a severe reaction given a food allergy). So, the probability is 0.07 * 0.41 = 0.0287, which can be rounded to 0.0287 or approximately 0.029.
To find the probability that a child younger than 18 is allergic to multiple foods, we use the information that 30% of those with an allergy are allergic to multiple foods. So, the probability is 0.07 (the probability of having a food allergy) multiplied by 0.30 (the probability of being allergic to multiple foods given a food allergy). Hence, the probability is 0.07 * 0.30 = 0.021, which can be rounded to 0.021 or approximately 0.021.
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The days maturity for a sample of 5 money market funds areshown here. The dollar amounts invested in the funds areprovided. Use the weighted mean to determine the mean numberof days to maturity for dollars invested in these 5 money marketfunds.COL1 Days tomaturity 20 12 7 5 6COL2 $$ Value (millions) 20 30 10 15 10
Final answer:
The weighted mean number of days to maturity for dollars invested in the 5 money market funds is approximately 11.35 days. This is calculated by taking the product of the days to maturity and the corresponding money value for each fund, summing these products, and then dividing by the total money value invested.
Explanation:
The question asks to calculate the weighted mean of days to maturity for dollars invested in several money market funds with varying maturities and dollar values. To compute this, we multiply each fund's days to maturity by its dollar value (in millions), sum these products, and then divide by the total of the dollar values. Here's the calculation:
(20 days * $20 million) + (12 days * $30 million) + (7 days * $10 million) + (5 days * $15 million) + (6 days * $10 million) = $400 million-days + $360 million-days + $70 million-days + $75 million-days + $60 million-days
Total million-days = $965 million-days
Total value of all funds = $85 million
Weighted mean days to maturity = Total million-days / Total value of all funds = $965 million-days / $85 million = 11.35 days
So, the weighted mean number of days to maturity for the dollars invested in these 5 money market funds is approximately 11.35 days.
The curve given by:
x=sin(????); y=sin(????+sin(????))
has two tangent lines at the point (x,y)=(0,0).
List both of them in order of increasing slope. Your answers should be in the form of y=????(x) without ????′????.
Answer:
Equations of tangent lines are
y= 2 x
y = 0
Step-by-step explanation:
x = sin t -- (1)
y = sin(t + sin(t)) -- (2)
Differentiating both equations w.r.to t to find slopes.
[tex]\frac{dx}{dt}=\frac{d(sin(t))}{dt}\\\\\frac{dx}{dt}=cos(t)--(3)[/tex]
[tex]\frac{dy}{dt}=\frac{d}{dt}(sin(t+sin(t))\\\\\frac{dy}{dt}=cos(t+sin(t))\frac{d}{dt}(t+sin(t))\\\\\frac{dy}{dt}=cos(t+sin(t)(1+cos(t))\\\\\frac{dy}{dt}=(1+cos(t))cos(t+sin(t))--(4)[/tex]
Dividing (2) by (1) to find slope
[tex]\frac{dy}{dx}=\frac{(1+cos(t))cos(t+sin(t))}{cos(t)}\\[/tex]
at tangent point x=y=0
From (1)
sin (t) = 0
⇒ t = 0, π
At t = 0
[tex]\frac{dy}{dx}\Big|_{t=0}=\frac{(1+cos(t))cos(t+sin(t))}{cos(t)}\\\\\\\frac{dy}{dx}\Big|_{t=0}=\frac{(1+cos(0))cos(0+sin(0))}{cos(0)}\\\\\\\frac{dy}{dx}\Big|_{t=0}=\frac{(1+1)cos(0+0)}{1}\\\\\\\frac{dy}{dx}\Big|_{t=0}=2\\[/tex]
At t= π
[tex]\frac{dy}{dx}\Big|_{t=\pi}=\frac{(1+cos(t))cos(t+sin(t))}{cos(t)}\\\\\\\frac{dy}{dx}\Big|_{t=\pi}=\frac{(1+cos(\pi))cos(\pi+sin(\pi))}{cos(\pi)}\\\\\\\frac{dy}{dx}\Big|_{t=\pi}=\frac{(1-1)cos(\pi+0)}{-1}\\\\\\\frac{dy}{dx}\Big|_{t=\pi}=0\\[/tex]
Equation of tangent
[tex](y-y_o)=m_t(x-x_o)\\[/tex]
[tex]Tangent\,\,point=(x_o,y_o)=(0,0)\\\\For\,\,t=0\\\\(y-0)=(2)(x-0)\\\\y=2x\\\\for\,\,t=\pi\\\\(y-0)=(0)(x-0)\\\\y=0[/tex]
Investments Suppose that you have $4000 to invest and you invest x dollars at 10% and the remainder at 896, write expressions in x that represent (a) the amount invested at 8%, (b) the interest earned on the x dollars at 10%, (c) the interest earned on the money invested at 8% (d) the total interest earned.
Answer:
Step-by-step explanation:
you have $4000 to invest and you invest x dollars at 10% and the remainder at 8℅.
a) an expression in x that represent the amount invested at 8% is
4000 - x
b) The The formula for simple interest is expressed as
I = PRT/100
Where
P represents the principal
R represents interest rate
T represents time in years
I = interest after t years
From the information given
P = $x
R = 10%
Assuming the investment is for 1 year, then interest,
I = (x × 10 × 1)/100
I = $0.1x
c) P = 4000 - x
R = 8℅
I = [(4000 - x) × 8 × 1)]/100
I = (32000 - 8x)/100
I = 320 - 0.08x
d) the total interest earned is
I = 0.1x + 320 - 0.08x
I = 0.02x + 320
A line has a slope of -3/5. Which ordered pairs could be points on a parallel line? Select two options.
The question is missing the options. The options are:
(A) (-8, 8) and (2, 2)
(B) (-5, -1) and (0, 2)
(C) (-3, 6) and (6.-9)
(D) (-2, 1) and (3,-2)
(E) (0, 2) and (5,5)
Answer:
Options (A) and (D)
Step-by-step explanation:
Given:
A line with slope (m) = [tex]-\frac{3}{5}[/tex]
Now, a parallel line to the given line will have the same slope.
So, let us check each of the given options.
Option (A)
(-8, 8) and (2, 2)
The slope of line passing through two points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given as:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Now, the slope of a line passing through (-8, 8) and (2, 2) is given as:
[tex]m_1=\frac{2-8}{2-(-8)}=\frac{-6}{10}=-\frac{3}{5}[/tex]
So, [tex]m=m_1[/tex]
Therefore, option (A) is correct.
Option (B): (-5, -1) and (0, 2)
The slope of a line passing through (-5, -1) and (0, 2) is given as:
[tex]m_2=\frac{2-(-1)}{0-(-5)}=\frac{3}{5}[/tex]
So, [tex]m\ne m_2[/tex]
Therefore, option (B) is not correct.
Option (C): (-3, 6) and (6, -9)
The slope of a line passing through (-3, 6) and (6, -9) is given as:
[tex]m_3=\frac{-9-6}{6-(-3)}=\frac{-15}{9}=-\frac{5}{3}\ne m[/tex]
Therefore, option (C) is not correct.
Option (D): (-2, 1) and (3, -2)
The slope of a line passing through (-2, 1) and (3, -2) is given as:
[tex]m_4=\frac{-2-1}{3-(-2)}=-\frac{3}{5}=m[/tex]
Therefore, option (D) is correct.
Option (E): (0, 2) and (5, 5)
The slope of a line passing through (0, 2) and (5, 5) is given as:
[tex]m_5=\frac{5-2}{5-0}=\frac{3}{5}\ne m[/tex]
Therefore, option (E) is not correct.
Hence, only options (A) and (D) are correct.
In 1990 the Department of Natural Resources released 1000 splake (a crossbreed of fish) into a lake. In 1997 the population of splake in the lake was estimated to be 3000. Using the Malthusian law for population growth, estimate the population of splake in the lake in the year 2020.
Using the Malthusian law for population growth and the given data, the estimated population of splake in a lake in 2020 is approximately 21,485.
Explanation:The question involves estimating the population of splake in a lake in 2020 using the Malthusian law for population growth. The Malthusian law indicates that populations grow exponentially under ideal conditions. Given that the population increased from 1000 to 3000 splake between 1990 and 1997, we can calculate the rate of growth and then apply this rate to predict the population in 2020.
To begin, we identify the years of growth as 1997 - 1990 = 7 years. The formula for exponential growth is P = P0ert, where P is the final population, P0 is the initial population, r is the rate of growth, and t is the time in years. With P = 3000, P0 = 1000, and t = 7, we can solve for r.
3000 = 1000e7r, which simplifies to 3 = e7r. Taking the natural logarithm of both sides gives us ln(3) = 7r, and solving for r gives r ~ 0.1487. Now, to find the population in 2020, which is 30 years from 1990, we use the formula with P0 = 1000, r = 0.1487, and t = 30: P = 1000e0.1487*30.
Upon calculation, the predicted population of splake in the year 2020 is approximately 21,485.
Shaki makes and sells backpack danglies. The total cost in dollars for Shaki to make q danglies is given by c(q)= 75+2q+0.015q^2 . Find the quantity that minimizes Shaki
the quantity that minimizes Shaki's cost is [tex]\( \frac{200}{3} \)[/tex], or approximately [tex]\( 66.67 \)[/tex] danglies.
To find the quantity that minimizes Shaki's cost function [tex]\( c(q) = 75 + 2q + 0.015q^2 \)[/tex], we need to find the value of q where the derivative of [tex]\( c(q) \)[/tex] with respect to [tex]\( q \)[/tex] is zero.
Given the cost function:
[tex]\[ c(q) = 75 + 2q + 0.015q^2 \][/tex]
We'll find the derivative [tex]\( c'(q) \)[/tex] with respect to q and set it equal to zero to find the critical points.
[tex]\[ c'(q) = \frac{d}{dq} (75 + 2q + 0.015q^2) \][/tex]
[tex]\[ c'(q) = 2 + 0.03q \][/tex]
Now, we'll set [tex]\( c'(q) \)[/tex] equal to zero and solve for q:
[tex]\[ 2 + 0.03q = 0 \][/tex]
[tex]\[ 0.03q = -2 \][/tex]
[tex]\[ q = \frac{-2}{0.03} \][/tex]
[tex]\[ q = -\frac{200}{3} \][/tex]
Since the quantity q must be positive in this context, we disregard the negative solution. Therefore, the critical point occurs at [tex]\( q = \frac{200}{3} \)[/tex].
To determine whether this critical point corresponds to a minimum, we'll analyze the second derivative [tex]\( c''(q) \)[/tex]. If [tex]\( c''(q) > 0 \)[/tex] at [tex]\( q = \frac{200}{3} \)[/tex], then it's a local minimum.
[tex]\[ c''(q) = \frac{d^2}{dq^2} (2 + 0.03q) \][/tex]
[tex]\[ c''(q) = 0.03 \][/tex]
Since [tex]\( c''(q) \)[/tex] is positive, the critical point [tex]\( q = \frac{200}{3} \)[/tex] corresponds to a minimum.
Therefore, the quantity that minimizes Shaki's cost is [tex]\( \frac{200}{3} \)[/tex], or approximately [tex]\( 66.67 \)[/tex] danglies.
Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). A new version of the exam was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 the mean score was 500.0 with a standard deviation of 10.6 . Use Table A to find the answers to the two questions.
(a) What proportion of students taking the MCAT had a score over 519 ?
(b) Compute the proportion and then enter the answer as a percentage rounded to two decimal places.
Answer:
(a) The proportion of students taking the MCAT had a score over 519 is 0.0367.
(b) The percentage of students who scored more than 519 in the MCAT is 3.67%.
Step-by-step explanation:
Assuming that the sample size or the number of students taking the MCAT in 2015 is large, the sampling distribution of scores follow a normal distribution.
Let X = score of a student
Given:
Mean = [tex]\mu=500[/tex]
Standard deviation = [tex]\sigma=10.6[/tex]
(a)
Compute the probability of students who scored more than 519 as follows:
[tex]P(X>519)=P(\frac{X-\mu}{\sigma}> \frac{519-500}{10.6})\\=P(Z>1.7925)\\=1-P(Z<1.7925)[/tex]
Use the z-table to determine the probability.
[tex]P(X>519)=P(\frac{X-\mu}{\sigma}> \frac{519-500}{10.6})\\=P(Z>1.7925)\\=1-P(Z<1.7925)\\=1-0.9633\\=0.0367[/tex]
Thus, the probability of students who scored more than 519 is 0.0367.
(b)
Convert the probability of students who scored more than 519 to percentage
[tex]=0.0367\times100\\=3.67\%[/tex]
Thus, the percentage of students who scored more than 519 is 3.67%.
Final answer:
The z-score for 519 is approximately 1.79, which correlates to about 3.67% of students scoring higher than 519.
Explanation:
To find the proportion of students scoring over 519 on the MCAT, first, we must calculate the z-score. The z-score tells us how many standard deviations an element is from the mean. Since the mean score is 500.0 and the standard deviation is 10.6, the z-score formula is z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. For a score of 519, the z-score would be z = (519 - 500.0) / 10.6 ≈ 1.79. To find the proportion of students scoring above this z-score, we would look up the z-score in a standard normal distribution table (Table A), or use a statistical software to find that the area to the right of z=1.79. This area corresponds to the proportion of students scoring higher than 519. Since standard normal distribution tables are commonly used and we don't have one provided here, let's assume that the area to the right of z=1.79 is approximately 3.67%.
g If there are 52 cards in a deck with four suits (hearts, clubs, diamonds, and spades), how many ways can you select 5 diamonds and 3 clubs?
Answer:
The number of ways to select 5 diamonds and 3 clubs is 368,082.
Step-by-step explanation:
In a standard deck of 52 cards there are 4 suits each consisting of 13 cards.
Compute the probability of selecting 5 diamonds and 3 clubs as follows:
The number of ways of selecting 0 cards from 13 hearts is:
[tex]{13\choose 0}=\frac{13!}{0!\times(13-0)!} =\frac{13!}{13!}=1[/tex]
The number of ways of selecting 3 cards from 13 clubs is:
[tex]{13\choose 3}=\frac{13!}{3!\times(13-3)!} =\frac{13!}{13!\times10!}=286[/tex]
The number of ways of selecting 5 cards from 13 diamonds is:
[tex]{13\choose 5}=\frac{13!}{5!\times(13-5)!} =\frac{13!}{13!\times8!}=1287[/tex]
The number of ways of selecting 0 cards from 13 spades is:
[tex]{13\choose 0}=\frac{13!}{0!\times(13-0)!} =\frac{13!}{13!}=1[/tex]
Compute the number of ways to select 5 diamonds and 3 clubs as:
[tex]{13\choose0}\times{13\choose3}\times{13\choose5}\times{13\choose0} = 1\times286\times1287\times1=368082[/tex]
Thus, the number of ways to select 5 diamonds and 3 clubs is 368,082.
How many names and binary predicates would a language like the first need in order to say everything you can say in the second?
Answer:The same number of names and 4 predicates
Step-by-step explanation:
Answer:
The same number of names and 4 predicates
Step-by-step explanation:
A campus deli serves 300 customers over its busy lunch period from 11:30 a.m. to 1:30 p.m. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?
Answer:
4 minutes
Step-by-step explanation:
There are two hours from 11:30 a.m. to 1:30 p.m
The hourly rate of service is:
[tex]r=\frac{300}{2}=150\ customers/hour[/tex]
If the average number of customers in the system (n) is 10, the time that a customer spends in process is given by:
[tex]t=\frac{n}{r} =\frac{10}{150}=0.06667\ hours[/tex]
Converting it to minutes:
[tex]t= 0.066667\ hours*\frac{60\ minutes}{1\ hour}\\t=4\ minutes[/tex]
A customer spends, on average, 4 minutes in process.
The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch. A. What is the probability that the diameter of a dot exceeds 0.0026 inch? B. What is the probability that a diameter is between 0.0014 and 0.0026? C. What standard deviation of diameters is needed so that the probability in part (b) is 0.995?
Answer:
(a) 0.06681
(b) 0.86638
(c) [tex]\sigma[/tex] = 0.000214
Step-by-step explanation:
We are given that the diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch i.e.; [tex]\mu[/tex] = 0.002 inch and [tex]\sigma[/tex] = 0.0004
Also, Z = [tex]\frac{X -\mu}{\sigma}[/tex] ~ N(0,1)
(a) Let X = diameter of a dot
P(X > 0.0026 inch) = P( [tex]\frac{X -\mu}{\sigma}[/tex] > [tex]\frac{0.0026 -0.002}{0.0004}[/tex] ) = P(Z > 1.5) = 1 - P(Z <= 1.5)
= 1 - 0.93319 = 0.06681
(b) P(0.0014 < X < 0.0026) = P(X < 0.0026) - P(X <= 0.0014)
P(X < 0.0026) = P( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{0.0026 -0.002}{0.0004}[/tex] ) = P(Z < 1.5) = 0.93319
P(X <= 0.0014) = P( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{0.0014 -0.002}{0.0004}[/tex] ) = P(Z <= -1.5) = 1 - P(Z <= 1.5)
= 1 - 0.93319 = 0.06681
Therefore, P(0.0014 < X < 0.0026) = 0.93319 - 0.06681 = 0.86638 .
(c) P(0.0014 < X < 0.0026) = 0.995
P( [tex]\frac{0.0014 -0.002}{\sigma}[/tex] < [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{0.0026 -0.002}{\sigma}[/tex] ) = 0.995
P( [tex]\frac{ -0.0006}{\sigma}[/tex] < Z < [tex]\frac{0.0006}{\sigma}[/tex] ) = 0.995
P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - P(Z <= [tex]\frac{-0.0006}{\sigma}[/tex] ) = 0.995
P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - (1 - P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) ) = 0.995
2 * P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - 1 = 0.995
P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) = 0.9975
On seeing the z table we observe that at critical value of x = 2.81 we get the probability area of 0.9975 i.e.;
[tex]\frac{0.0006}{\sigma}[/tex] = 2.81 ⇒ [tex]\sigma[/tex] = 0.000214
Therefore, 0.000214 standard deviation of diameters is needed so that the probability in part (b) is 0.995 .
The probabilities of obtaining a given diameter is found from the z-table,
given that the dot produced by the printer are normally distributed.
A. The probability that the diameter of a dot exceeds 0.0026 inch is 0.0668B. The probability that the diameter is between 0.0014 and 0.0026 inch is 0.8664C. The standard deviation needed for a probability of 0.995 is 2.135 × 10⁻⁴Reasons:
The mean diameter, μ = 0.002
The standard deviation, σ = 0.0004
A. The probability that the diameter exceeds 0.0026 inch
Solution;
[tex]\displaystyle z-score,\ Z= \mathbf{\dfrac{x-\mu }{\sigma }}[/tex]
At x = 0.0026 inch, we have;
[tex]\displaystyle Z=\dfrac{0.0026-0.002 }{0.0004 } = 1.5[/tex]
P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668
The probability that the diameter of a dot exceeds 0.0026 inch = 0.0668
B. The probability that the diameter is less than 0.0026 inch = 0.9332
The z-score for a diameter of x = 0.0014 inch is given as follows;
[tex]\displaystyle Z=\mathbf{\dfrac{0.0014-0.002 }{0.0004 }} = -1.5[/tex]
P(Z < -1.5) = 0.0668
Therefore, the probability that the diameter is between 0.0014 and 0.0026 inch is given as follows;
P(0.0014 < x < 0.0026) = 0.9332 - 0.0668 = 0.8664
The probability that the diameter is between 0.0014 and 0.0026 = 0.8664
C. For the probability in part (b) to be 0.995, we have;
For a probability of 0.995, the z-score ≈ 2.575
[tex]\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)- P\left(Z < \dfrac{0.0014-0.002 }{\sigma } \right)= 0.995[/tex]
Therefore;
[tex]\displaystyle \mathbf{ P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)} = 0.995 + \frac{1 - 0.995}{2} = 0.9975[/tex]
From the z-table, we get;
[tex]\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right) = 0.9975[/tex]
The z-score with a probability of 0.9975 = 2.81
Which gives;
[tex]\displaystyle \left( \dfrac{0.0026-0.002 }{\sigma } \right) = 2.81[/tex]
[tex]\displaystyle \sigma = \left( \dfrac{0.0026-0.002 }{2.81} \right) = \mathbf{2.135 \times 10^{-4}}[/tex]
The standard deviation of the diameters needed so that the probability in part (b) is 0.995 is 2.135 × 10⁻⁴
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An adult male African elephant weighs about 9.07*10^3 kg. Compute how many times heavier an adult male blue whale is than an adult male African elephant(I.e., find the value of the ratio). Round your final answer to the nearest tenth.
Answer:
An adult male blue whale is 18.7 times heavier than an adult male African elephant.
Step-by-step explanation:
As the weight of an adult male African elephant weighs about
[tex]9.07\:\times\:10^3[/tex] kgAnd the weight of an adult blue whale is
[tex]1.7\:\times\:10^5[/tex] kgDetermining the ratio of adult male African elephant to the weight of an adult blue whale as:
[tex]\:\:\frac{1.7\:\times\:10^5}{9.07\:\times\:10^3}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}[/tex]
[tex]=\frac{10^{5-3}\times \:1.7}{9.07}[/tex]
[tex]=\frac{10^2\times \:1.7}{9.07}[/tex]
As [tex]10^2\times \:1.7=170[/tex], So
[tex]=\frac{170}{9.07}[/tex]
[tex]=18.74310\dots[/tex]
Round the answer to the nearest tenth
[tex]=18.7[/tex]
Therefore, an adult male blue whale is 18.7 times heavier than an adult male African elephant.
Keywords: ratio, nearest tenth
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A researcher has a hypothesis that a specific drug may have a higher prevalence of side effects among members of the African American population than members of the Caucasian population. Which statistical technique might the researcher want to use when designing a study to test their hypothesis
A. Stratification
B. Crossover matching
C. Matching
D. Randomization
Answer:
A. Stratification
Step-by-step explanation:
Stratified random sampling is used when the researcher wants to highlight a specific subgroup within an entire population.
Stratification technique is mainly used to reduce the population differences and to increase the efficiency of the estimates. In this method the population is divided into a number of subgroups or strata.
Each strata should be so formed such that they are homogeneous as far as possible.
Final answer:
When examining the hypothesis about drug side effects in different populations, Randomization is the most appropriate statistical technique. It helps in reducing bias and ensures equal chances of group assignment, providing more reliable results in comparing the effects across populations.(Option D)
Explanation:
A researcher examining the hypothesis that a specific drug may exhibit a higher prevalence of side effects among the African American population compared to the Caucasian population could employ several statistical techniques to design the study. However, the most appropriate option provided is Randomization. Randomization helps in mitigating bias and ensures that each participant has an equal chance of being assigned to either the experimental or control group. This process decreases the likelihood of systematic differences between groups and allows any effects observed to be more confidently attributed to the drug under study rather than external factors.
Other options like Stratification, Crossover matching, and Matching could also play roles in different aspects of study design, but when testing a hypothesis about a drug's effects across different populations, randomization is crucial. It aligns with principles of experimental design that seek to control, to the extent possible, for variables that could influence the outcome, ensuring that the treatment group and control group are comparable at the beginning of the study.
The sum of the diameters of the largest and smallest pizzas sold at a pizza shop is 25 inches. The difference in their diameters is 15 inches. Find the diameters of the largest and smallest pizzas.
Answer:
20 inches and 5 inches
Step-by-step explanation:
Let the diameter of the largest and the smallest pizza be Y and X respectively.
Then,
Y+X = 25 ........................... Equation 1
Y-X = 15 ............................ Equation 2
Solve equation 1 and equation 2 simultaneously.
Add equation 1 and equation 2
Y+Y = +X+(-X)+25+15
2Y = 40
Y = 40/2
Y = 20 inches.
Also,
Substitute the value of Y into equation 1
20+X=25
X = 25-20
X = 5 inches.
Hence the diameter of the largest and the smallest pizzas = 20 inches and 5 inches
Final answer:
The smallest pizza has a diameter of 5 inches, and the largest pizza has a diameter of 20 inches.
Explanation:
The diameters of the largest and smallest pizzas are 20 inches and 5 inches, respectively.
To find the diameters of the pizzas:
Let x be the diameter of the smallest pizza and y be the diameter of the largest pizza.
We have the system of equations x + y = 25 and y - x = 15.
Solving these equations simultaneously, we get y = 20 and x = 5.
The list of digits below is from a random number generator using technology. Use the list of numbers to obtain a simple random sample of size 3 from this list. If you start on the left and take the first three numbers between 1 and 9, what three books would be selected from the numbered list?
Question Continuation
5 2 5 5 2 1 0 5 7 5 8 9 3 7 2
Options
A. A Tale of Two Cities, Huckleberry Finn, A Tale of Two Cities
B. A Tale of Two Cities, Huckleberry Finn, The Sun Also Rises
C. A Tale of Two Cities, Huckleberry Finn, Crime and Punishment
D. Huckleberry Finn, Crime and Punishment, The Jungle
E. Crime and Punishment, The Jungle, The Sun Also Rises
Book List
1. Crime and Punishment
2. Huckleberry Finn
3. The Sun Also Rises
4. As I Lay Dying
5. A Tale of Two Cities
6. Death of a Salesman
7. The Jungle
8. Pride and Prejudice
9. The Scarlet Letter
Answer:
C. A Tale of Two Cities, Huckleberry Finn, Crime and Punishment
Step by step explanation
Counting from the left, the selected numbers are 5 , 2 and 1
The books are
5. A Tale of two cities
2. Huckleberry Finn
1. Crime and Punishment
Note that the numbers on the list are 5 2 5 5 2 1
After book 5 and 2 have been selected, the next series of numbers (5 5 2) can not be considered because they've already been selected.
So, the next number after 5 2 5 5 2 is then selected, which is 1
The selected books are:
The books are: A Tale of two cities, Huckleberry Finn, Crime and Punishment
The simple random selection of three books using the random number generated will include the books : The Sun also rises, The Scarlet letter, Crime and Punishment.
The random number generated using technology include :
7, 2, 7, 2, 2, 6, 7, 0, 8, 3, 2, 8, 5, 3, 1
Making a selection of 3 numbers between (1 - 9) starting from the left hand side of the list : 7, 2, 6 ( repeated numbers are only chosen once) as we have to make a unique selection of numbers.From the list of number books attached below :
7. The Sun also rise
2. The Scarlet letter
6. Crime and Punishment
Hence, the randomly selected books will be :
The Sun also rises, The Scarlet letter, Crime and Punishment.
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Please help!!!!!!!!
Answer:
a) 62
b) 24
Step-by-step explanation:
For A, add the students who watched only one movie: 18+24+20=62
For B, look at how many students only watched Star Wars: 24
A box contains 11 two-inch screws, of which 4 have a Phillips head and 7 have a regular head. Suppose that you select 3 screws randomly from the box with replacement. Find the probability there will be more than one Phillips head screw.
Answer:
The probability that there will be more than one Phillips head screw = 0.1803 .
Step-by-step explanation:
We are given that there are 11 two-inch screws in a box of which 4 have a Phillips head and 7 have a regular head.
We are selecting 3 screws randomly from the box with replacement, so the probability that there will be more than one Phillips head screw is given by :
Probability of selecting two Phillips head screw.Probability of selecting three Phillips head screw.Now P(selecting 2 Phillips head screw with replacement) is given by :
Selecting 2 Phillip head screw = [tex]\frac{4}{11}[/tex] * [tex]\frac{4}{11}[/tex] = [tex]\frac{16}{121}[/tex]
P(selecting three Phillips head screw) = [tex]\frac{4}{11}[/tex] * [tex]\frac{4}{11}[/tex] * [tex]\frac{4}{11}[/tex] = [tex]\frac{64}{1331}[/tex]
Therefore, Probability that there will be more than one Phillips head screw
= [tex]\frac{16}{121}[/tex] + [tex]\frac{64}{1331}[/tex] = [tex]\frac{240}{1331}[/tex] = 0.1803 .
Set up the integral that uses the method of disks/washers to find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified lines.
y=3\sqrt(x), y = 3, x= 0
a.) about the line y = 3
b.) about the line x = 5
The integrals for the volume of the solids
a.) V = ∫[0 to 1] π(3 - 3√x)² dx
b.) V = ∫[0 to 27] π(5 - [tex]x^{2/3}[/tex]))² dx
We have,
To set up the integral using the method of disks/washers to find the volume V of the solid obtained by rotating the region bounded by the curves y = 3√x, y = 3, and x = 0 about the specified lines, follow these steps:
Given curves: y = 3√x, y = 3, x = 0
a.)
Rotating about the line y = 3:
- Draw the region bounded by the curves y = 3√x, y = 3, and x = 0.
- The solid will be formed by revolving this region around the line y = 3.
- For the method of disks/washers, consider a vertical slice (dx) of thickness dx at a distance x from the y-axis.
- The radius of the disk is the distance between the curve y = 3√x and the line y = 3, which is (3 - 3√x).
- The area of the disk is π(radius)^2 = π(3 - 3√x)².
- The volume of the infinitesimally thin disk is dV = π(3 - 3√x)² dx.
- Integrate the volume from x = 0 to x = (3/3)² = 1:
V = ∫[0 to 1] π(3 - 3√x)² dx
b)
Rotating about the line x = 5:
- Draw the region bounded by the curves y = 3√x, y = 3, and x = 0.
- The solid will be formed by revolving this region around the line x = 5.
- For the method of disks/washers, consider a vertical slice (dx) of thickness dx at a distance x from the y-axis.
- The radius of the disk is the distance between the line x = 5 and the curve y = 3√x, which is (5 - [tex]x^{2/3}[/tex]).
- The area of the disk is π(radius)^2 = π(5 - [tex]x^{2/3}[/tex])².
- The volume of the infinitesimally thin disk is dV = π(5 - [tex]x^{2/3}[/tex])² dx.
Integrate the volume from x = 0 to x = 3³ = 27:
V = ∫[0 to 27] π(5 - [tex]x^{2/3}[/tex])² dx
These integrals will give you the volumes of the solid obtained by rotating the region about the specified lines. You can evaluate these integrals to find the exact values of the volumes.
Thus,
The integrals for the volume of the solids
a.) V = ∫[0 to 1] π(3 - 3√x)² dx
b.) V = ∫[0 to 27] π(5 - [tex]x^{2/3}[/tex]))² dx
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According to CNN business partner Careerbuilder, the average starting salary for accounting graduates in 2018 was at least $57,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2018. State the appropriate null and alternative hypotheses.
Answer:
Null hypothesis: The American Society for Certified Public Accountants says the average starting salary of accountants who graduated in 2018 is $57,413
Alternate hypothesis: The American Society for Certified Public Accountants says the average starting salary of accountants who graduated in 2018 is less than or equal to $57,413
Step-by-step explanation:
A null hypothesis is a statement from a population parameter that is subject to testing. It is expressed with equality.
An alternate hypothesis is also a statement from the population parameter that negates the null hypothesis. It is expressed with inequality
A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 35 cm/s.
Express the radius r of this circle as a function of the time t (in seconds).
Answer:
The radius r of this circle as a function of the time t :
[tex]r(t)=35\times t[/tex]
Step-by-step explanation:
Speed of the circular ripple = S = 35 cm/s
Radius of the ripple at time t = r
[tex]Speed=\frac{Distance}{Time} [/tex]
[tex]35cm/s=\frac{r}{t}[/tex]
[tex]r=35 cm/s \times t[/tex]
The radius r of this circle as a function of the time t :
[tex]r(t)=35\times t[/tex]
The radius r of this circle is a function of the time t (in seconds) is 35t.
Given that
A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 35 cm/s.
We have to determineThe radius r of this circle is a function of the time t (in seconds).
According to the questionSpeed of the circular ripple = S = 35 cm/s
Radius of the ripple at time t = r
Then
The radius r of this circle is a function of the time t (in seconds) is determined by the following formula;
[tex]\rm Speed = \dfrac{Distance}{Time}[/tex]
Substitute all the values in the formula;
[tex]\rm Speed = \dfrac{Distance}{Time}\\ \\ 35 = \dfrac{r}{t}\\ \\ r = 35 \times t\\ \\ r = 35t[/tex]
Hence, The radius r of this circle is a function of the time t (in seconds) is 35t.
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The following six independent length measurements were made (in feet) for a line: 736.352, 736.363, 736.375, 736.324, 736.358, and 736.383. Determine the standard deviation of the measurements.
Answer:
Assuming population data
[tex] \sigma = \sqrt{0.000354}=0.0188[/tex]
Assuming sample data
[tex] s = \sqrt{0.000425}=0.0206[/tex]
Step-by-step explanation:
For this case we have the following data given:
736.352, 736.363, 736.375, 736.324, 736.358, and 736.383.
The first step in order to calculate the standard deviation is calculate the mean.
Assuming population data
[tex]\mu = \frac{\sum_{i=1}^6 X_i}{6}[/tex]
The value for the mean would be:
[tex]\mu = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359[/tex]
And the population variance would be given by:
[tex] \sigma^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6}[/tex]
And we got [tex] \sigma^2 =0.000354[/tex]
And the deviation would be just the square root of the variance:
[tex] \sigma = \sqrt{0.000354}=0.0188[/tex]
Assuming sample data
[tex]\bar X = \frac{\sum_{i=1}^6 X_i}{6}[/tex]
The value for the mean would be:
[tex]\bar X = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359[/tex]
And the population variance would be given by:
[tex] s^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6-1}[/tex]
And we got [tex] s^2 =0.000425[/tex]
And the deviation would be just the square root of the variance:
[tex] s = \sqrt{0.000425}=0.0206[/tex]
To determine the standard deviation of given length measurements, calculate the mean, find squared differences, average them, and take the square root. The standard deviation of the measurements is 0.01879 feet.
Determining the Standard Deviation of Measurements
To find the standard deviation of the given length measurements, follow these steps:
Calculate the mean of the measurements.
Find the squared differences between each measurement and the mean.
Compute the average of the squared differences.
Take the square root of that average.
Step-by-Step Calculation
→ List of measurements: 736.352, 736.363, 736.375, 736.324, 736.358, 736.383.
→ Calculate the mean:
= (736.352 + 736.363 + 736.375 + 736.324 + 736.358 + 736.383) / 6
= 736.3591667
Find the squared differences:
(736.352 - 736.3591667)^2 = 0.00005256(736.363 - 736.3591667)^2 = 0.00001464(736.375 - 736.3591667)^2 = 0.00024811(736.324 - 736.3591667)^2 = 0.00123264(736.358 - 736.3591667)^2 = 0.00000136(736.383 - 736.3591667)^2 = 0.00057044→ Calculate the average of the squared differences:
= (0.00005256 + 0.00001464 + 0.00024811 + 0.00123264 + 0.00000136 + 0.00057044) / 6
= 0.000353292
Take the square root to find the standard deviation:
= √(0.000353292)
= 0.01879 feet
Which, if any, of A. (4, π/6), B. (−4, 7π/6), C. (4, 13π/6), are polar coordinates for the point given in Cartesian coordinates by P(2, 2 √ 3)?
Final answer:
Explaining the polar coordinates for given Cartesian coordinates.
Explanation:
Polar Coordinates of Points:
Point A(2, 2√3): Polar coordinates are (4, π/6).
Point B(-4, 7π/6): Wrong polar coordinates as it should be (4, 11π/6).
Point C(4, 13π/6): Wrong polar coordinates.
f left parenthesis x right parenthesis equals StartFraction 16 x squared Over x Superscript 4 Baseline plus 64 EndFractionf(x)=16x2 x4+64(a) Is the pointleft parenthesis negative 2 StartRoot 2 EndRoot comma 1 right parenthesis−22,1on the graph of f?(b) Ifx equals 2 commax=2,what is f(x)? What point is on the graph of f?(c) If f left parenthesis x right parenthesis equals 1 commaf(x)=1, what is x? What point(s) is (are) on the graph of f?(d) What is the domain of f?(e) List the x-intercepts, if any, of the graph of f.(f) List the y-intercept, if there is one, of the graph of f.
Answer:
a) Yes
b) (2,0.8)
c)[tex](2\sqrt2,1), (-2\sqrt2,1)[/tex]
d) [tex]x \in (-\infty,\infty)[/tex]
e) (0,0)
f) (0,0)
Step-by-step explanation:
We are given the following function in the question:
[tex]f(x) = \displaystyle\frac{16x^2}{x^4 + 64}[/tex]
a) We have to check whether given point lies on the function or not.
[tex](-2\sqrt2,1)\\\\f(-2\sqrt2) = \displaystyle\frac{16(-2\sqrt2)^2}{(-2\sqrt2)^4 + 64} = \frac{128}{128} = 1[/tex]
b) Find value of f(x) at x = 2
[tex]f(2) = \displaystyle\frac{16(2)^2}{(2)^4 + 64} =\frac{64}{80}= 0.8[/tex]
Thus, (2,0.8) lies on the graph of given function.
c) We have to find the value of x, when f(x) = 1
[tex]1 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\x^4 -16x^2 + 64 = 0\\(x^2-8)^2 = 0\\x^2 - 8 = 0\\x = \pm 2\sqrt2[/tex]
[tex](2\sqrt2,1), (-2\sqrt2,1)[/tex] lies on he graph of function.
d) Domain is the collection of all values of x for which the function is defined.
[tex]x \in (-\infty,\infty)[/tex]
e) x-intercepts
This is the value of x such that the function is zero.
[tex]0 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\16x^2 = 0\\x = 0[/tex]
f) y-intercept
It is the value of function when x is zero.
[tex]f(0) = \displaystyle\frac{16(0)^2}{(0)^4 + 64} = 0[/tex]
The function passes trough origin.
The solution involves determining if a specific coordinates exist on the graph, calculating the function value for specific x-values, determining x-values for a specific function value, finding the domain of the function, and identifying the x and y intercepts of the function.
Explanation:The function is f(x) = 16x²/(x⁴ + 64).
To check if the point (-2√2, 1) is on the graph, substitute x = -2√2 into f(x). If f(-2√2) equals 1, the point is on the graph.For x = 2, substitute x = 2 into f(x) to get f(2). The point on the graph for this x-value will be (2, f(2)).To find x when f(x) = 1, set f(x) equal to 1 and solve for x. The points on the graph will be (x, 1), where x are the solutions to the equation.The domain of the function f is all real numbers except for x-values that make the denominator zero. Solve x⁴ + 64 = 0 to find x-values to exclude from the domain.To find the x-intercepts of the graph, set f(x) equal to zero and solve for x.The y-intercept of the graph is the value of f(x) at x = 0, which is f(0).Learn more about Function Analysis here:https://brainly.com/question/34156091
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The amount of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified at possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of 0.2 ounces (these are the population parameters). Suppose a sample of 100 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 100 bags is less than 10.45 ounces. (Hint: think of this in terms of a sampling distribution with sample size
Answer:
0.62% probability that the sample mean weight of these 100 bags is less than 10.45 ounces.
Step-by-step explanation:
To solve this question, the concepts of the normal probability distribution and the central limit theorem are important.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 10.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02[/tex]
Find the probability that the sample mean weight of these 100 bags is less than 10.45 ounces
This is the pvalue of Z when X = 10.45. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.45 - 10.5}{0.02}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062.
So there is a 0.62% probability that the sample mean weight of these 100 bags is less than 10.45 ounces.
The probability of the sample mean weight being less than 10.45 ounces can be found by calculating the Z-score and referencing a standard normal distribution table. The calculated Z-score (-2.5) corresponds to a probability of approximately 0.62%.
Explanation:The problem is about determining the probability that the sample mean weight of corn chip bags is less than 10.45 ounces. This is a problem of finding a probability in a sampling distribution when the population parameters are known. Given the data, we can use the Central Limit Theorem, which states that if the sample size is large enough (usually >30), the sampling distribution approximates a normal distribution.
To solve this, you can use the formula Z = (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Plugging in the given values: Z = (10.45 - 10.5) / (0.2 / √100) = -2.5. The Z-score tells us how many standard deviations away our data point is from the mean. To find the probability that the Z is less than -2.5, you can refer to a standard normal distribution table or use statistical software. According to the Z table, the probability is approximately 0.0062 or 0.62% that the sample mean weight of these 100 bags is less than 10.45 ounces.
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b=2.35 + 0.25x
c=1.75+0.40x
In the equations above,b and c represent the price per pound,in dollars, of beef and chicken,respectively,x weeks after July 1 of last summer.What was the price per pound of beef when it was equal to the price per pound of chicken?
A.2.60
B.2.85
C.2.95
D.3.35
Answer:
D. 3.35
Step-by-step explanation:
First we need to form an equation and solve it to find the number of weeks when the prices were the same. Because the prices were the same we can say that b = c, and therefore form the equation:
2.35 + 0.25x = 1.75 + 0.4x - Now we nee to solve it and find x.
2.35 - 1.75 = 0.4x - 0.25x
0.6 = 0.15x
x = 0.6 ÷ 0.15
x = 4 weeks
So now we substitute x into the equation for beef and find the price.
b = 2.35 + (0.25 × 4)
b = 2.35 + 1
b = $3.35 per pound
The price per pound of beef when it was equal to the price of chicken is $3.35 per pound.
The prices of beef and chicken are represented by the following linear equations :
Price of beef, B :
B = 2.35 + 0.25x
Price of chicken, C :
C = 1.75 + 0.40x
Where x in both equations represents x weeks after July 1 of last summer :
Firstly :
We find the week in which the price of beef and chicken are the same :
Beef = Chicken
2.35 + 0.25x = 1.75 + 0.40x
We solve for x
2.35 - 1.75 = 0.40x - 0.25x
0.60 = 0.15x
x = 0.60 / 0.15
x = 4
Therefore, 4 weeks after July 1 of last summer, the price of beef and chicken were the same.
Therefore, the price per pound of beef in the 4th week is :
B = 2.35 + 0.25(4)
B = 2.35 + 1
B = 3.35
The price per pound of beef when it was equal to the price of chicken is $3.35 per pound.
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Explain what a P-value is. What is the criterion for rejecting the null hypothesis using the P-value approach?
Answer:
P-value or Probability value is the exact percentage where test statistics lie.The criterion for rejecting the null hypothesis using the P-value approach is that if P-value < Level of significance , then we reject our null hypothesis
Step-by-step explanation:
P-value or Probability value is the exact percentage where test statistics lie.
It also tells the probability of obtaining extreme results corresponding to our level of significance keeping in state that our null hypothesis is true or correct.
The criterion for rejecting the null hypothesis using the P-value approach is that if P-value < Level of significance , then we reject our null hypothesis i.e.
Suppose P-value is 2.33% and Level of significance is 5%, then we will reject our null hypothesis as 2.33% < 5%.
On the other hand, if P-value > Level of significance , then we cannot reject or accept our null hypothesis.
A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level.
Explanation:A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. It represents the probability of observing a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true. In other words, it quantifies how likely it is for the observed data to have occurred by chance if the null hypothesis is true.
The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level (usually denoted as alpha). If the p-value is smaller than alpha, we reject the null hypothesis, indicating that there is enough evidence to support the alternative hypothesis. Alternatively, if the p-value is greater than or equal to alpha, we fail to reject the null hypothesis, suggesting that there is not enough evidence to support the alternative hypothesis.