Answer:
The height of the missing rectangle is 0.15Explanation:
The image attached has the mentioned histogram.
Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.
Only the rectangle for the class [3,4] is missing.
The height of each rectangle is the relative frequency of the corresponding class.
The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.
In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.
1. Sum of the known relative frequencies:
0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.852. Missing frequency:
1 - 0.85 = 0.153. Conclusion:
The height of the missing rectangle is 0.15How does hypothesis testing differ from constructing confidence intervals, in general? Read carefully.
Answer:
Step-by-step explanation:
Hypothesis testing invariably used to test a claim about a population parameter is widely used to check whether they hypothetical claim made is right.
For example, mean scores of a particular college is more than 75% is tested with hypothesis as setting null as equal to 75% and alternate >75%
Processes are done stepwise from the sample collected and conclusion made
Confidence interval on the other hand is the range of values within which the parameter is expected to lie at a certain confidence level
Estimated population parameter is provided for error known as margin of error depending upon the confidence level, and an interval is prepared which guarantees to the extent of confidence that parameters will fall within.
Hypothesis testing can be concluded with the use of confidence intervals also.
The average number of customers served by The Copy Shop during a typical morning (9am to noon) is 12. One morning, The Copy Shop has to close for 15 minutes.
What is the probability that no customers will arrive during this 15 minute period?
X = number of customers
a. X ~ binomial
b. X ~ negative binomial
c. X ~ hypergeometric
d. X ~ Poisson
Answer: d, p = 0.4493
Step-by-step explanation: this question is solved using a possion probability distribution because the event is occurring at a fixed rate.
For this question of ours the fixed rate is the fact that 12 customers visiting the shop within 15 minutes.
For this question our fixed rate (u) = 12/15 = 0.8
The probability distribution for possion is given as
P(x=r) = (e^-u * u^r) / r!
At this point x = 0 ( no customers coming to the shop)
P(x=0) =( e^-0.8 * 0.8^0)/ 0!
P(x=0) = (e^-0.8 * 1)/1
P(x=0) = e^-0.8
P(x=0) = 0.4493
The following probability distributions of jobsatisfaction scores for a sample of informationsystems (IS) senior executives and IS middle managersrange from a low of 1 (very dissatisfied) to a high of5 (very satisfied).Probability Job Satisfaction Score IS SeniorExecutives 1 .05 2 .093 .03 4 .425 .41IS Middle Managers.04.10.12.46.28a. What is the expected value of the job satisfactionscore for senior executives?b. What is the expected value of the job satisfactionscore for middle managers?c. Compute the variance of job satisfaction scores forexecutives and middle managers.d. Compute the standard deviation of job satisfactionscores for both probability distributions.e. Compare the overall job satisfaction of seniorexecutives and middle managers.
Answer:
a) 4.076
b) 3.9
c) variance for executives=1.128
variance for middle mangers=0.73
d)standard deviation for executives=1.062
standard deviation for middle mangers=0.854
e) Overall job satisfaction for senior executives is higher than middle manager.
Step-by-step explanation:
IS senior executives
Job Satisfaction 1 2 3 4 5
Probability 0.05 0.093 0.03 0.425 0.41
IS middle manager
Job Satisfaction 1 2 3 4 5
Probability 0.01 0.1 0.12 0.46 0.28
Let X denotes IS senior executive and Y denotes IS middle manager.
a)
E(X)=∑x*p(x)=1*0.05+2*0.093+3*0.03+4*0.425+5*0.41
E(X)=0.05+0.186+0.09+1.7+2.05
E(X)=4.076
b)
E(Y)=∑y*p(y)=1*0.1+2*0.1+3*0.12+4*0.46+5*0.28
E(Y)=0.1+0.2+0.36+1.84+1.4
E(Y)=3.9
c)
V(x)=∑x²*p(x)-(∑x*p(x))²
∑x²*p(x)=1*0.05+4*0.093+9*0.03+16*0.425+25*0.41
∑x²*p(x)=0.05+0.372+0.27+6.8+10.25
∑x²*p(x)=17.742
V(x)=17.742-(4.076)²
V(x)=1.128
V(y)=∑y²*p(y)-(∑y*p(y))²
∑y²*p(y)=1*0.1+4*0.1+9*0.12+16*0.46+25*0.28
∑y²*p(y)=0.1+0.4+1.08+7.36+7
∑y²*p(y)=15.94
V(y)=15.94-(3.9)²
V(y)=0.73
d)
S.D(x)=√V(x)
S.D(x)=√1.128
S.D(x)=1.062
S.D(y)=√V(y)
S.D(y)=0.854
e)
Overall job satisfaction for senior executives is more than middle manager as expected value of senior executives is greater than expected value of middle manger with relatively higher variability than middle manager.
There are several ways you might think you could enter numbers in WebAssign, that would not be interpreted as numbers. N.B. There may be hints in RED!!!
-You cannot have commas in numbers.
-You cannot have a space in a number.
-You cannot substitute the letter O for zero or the letter l for 1.
-You cannot include the units or a dollar sign in the number.
-You can have the sign of the number, + or -.
Which of the entries below will be interpreted as numbers?
a) 1.56e-9
b) -4.99
c) 40O0
d) 1.9435
e) 1.56 e-9
f) $2.59
g) 3.25E4
h) 5,000
i) 1.23 inches
The entries that will be interpreted as numbers are a) 1.56e-9, b) -4.99, d) 1.9435, and g) 3.25E4. Entries c), e), f), h), and i) violate the restrictions mentioned.
Explanation:The entries that will be interpreted as numbers are:
a) 1.56e-9b) -4.99d) 1.9435g) 3.25E4Entries a), b), d), and g) are written in proper scientific notation and do not violate any of the restrictions mentioned. Entry c) violates the rule of substituting the letter O for zero or the letter l for 1. Entry e) has a space between the number and the exponent, violating the rule of not having a space in a number. Entry f) includes a dollar sign in the number, which is not allowed. Entry h) has a comma, violating the rule of not having commas in numbers. Entry i) includes the units 'inches', which is not allowed.
Final answer:
Entries 1.56[tex]e^{-9[/tex], -4.99, and 1.9435 are valid numbers that would be interpreted by WebAssign, while 3.25E4 is also correct as it is proper scientific notation. (Option a, b,d,g)
Explanation:
When entering numbers in WebAssign, it is important to use the correct format to ensure that the numbers are interpreted accurately. Here are which of the entries will be interpreted as numbers:
a) 1.56[tex]e^{-9[/tex]: This is a correct representation of a number in scientific notation.
b) -4.99: This is a valid negative number.
c) 40O0: This contains a letter and would not be interpreted as a number.
d) 1.9435: This is a simple decimal number.
e) 1.56 [tex]e^{-9[/tex]: This is not correctly formatted for scientific notation due to the space.
f) $2.59: This includes a dollar sign, which is not valid for a number entry.
g) 3.25E4: This is correctly written in scientific notation.
h) 5,000: Commas are not allowed in number entries.
i) 1.23 inches: This includes units, which should not be included in a number entry.
A lock has two buttons: a "0" button and a "1" button.To open a door you need to push the buttons according to a preset 8-bit sequence. How many sequences are there? Suppose you press an arbitrary 8-bit sequence; what is the probability that the door opens? If the first try does not succeed in opening the door, you try another number; what is the probability of success?
Answer:
Step-by-step explanation:
The problem relates to filling 8 vacant positions by either 0 or 1
each position can be filled by 2 ways so no of permutation
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
b )
Probability of opening of lock in first arbitrary attempt
= 1 / 256
c ) If first fails , there are remaining 255 permutations , so
probability of opening the lock in second arbitrary attempt
= 1 / 255 .
Solve the following
4x + 2 y = 16
3x + 3y =15
answer :- y = 9/2
and x = 1/2
Answer:
Step-by-step explanation:
The given system of equations is expressed as
4x + 2y = 16 - - - - - - - - - - - - -1
3x + 3y =15- - - - - - - - - - - - - -2
We would eliminate x by multiplying equation 1 by 3 and equation 2 by 4. It becomes
12x + 6y = 48 - - - - - - - - - - - -3
12x + 12y = 60 - - - - - - - - - - - -4
Subtracting equation 4 from equation 3, it becomes
- 6y = - 12
Dividing the left hand side and the right hand side of the equation by - 6, it becomes
- 6y/-6 = - 12/ - 6
y = 2
Substituting y = 2 into equation 1, it becomes
4x + 2 × 2 = 16
4x + 4 = 16
Subtracting 4 from the left hand side and the right hand side of the equation, it becomes
4x + 4 - 4 = 16 - 4
4x = 12
Dividing the left hand side and the right hand side of the equation by 4, it becomes
4x/4 = 12/4
x = 3
A force of 10 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length?
Answer:
Work done will be equal to 5.2059 lb-ft
Step-by-step explanation:
We have given force F = 10 lb
Spring is stretched to 2 in
So x = 2 in
As 1 inch = 0.0833 feet
So 2 inch = 2×0.0833 = 0.1666 feet
From hook's law we know that F = Kx , here K is spring constant and x is spring elongation
So [tex]10=K\times 0.1666[/tex]
K = 60.024 lb/feet
Now new elongation x = 5 in
So 5 in = 5×0.0833 = 0.4165 feet
Work done is given by [tex]W=\frac{1}{2}Kx^2[/tex]
So [tex]W=\frac{1}{2}\times 60.02\times 0.4165^2=5.205lb-ft[/tex]
So work done will be equal to 5.2059 lb-ft
The Big Falcon Rocket (BFR or Starship) from Space X can carry approximately 220,000 pounds. If they only carried $100 bills, how much money can they carry?
Answer:
$9,979,032,100 or $9.979 billion
Step-by-step explanation:
The approximate weight of a $100 bill is 1 gram.
All of the calculations bellow assume that the volume of the bills would not be an issue and only concerns weight.
1 pound is equivalent to approximately 453.59237 grams.
The weight in grams that the Big Falcon Rocket can carry is:
[tex]W= 220,000*453.59237=99,790,321.4\ g[/tex]
Since each bill weighs 1 gram, the number of bills it could carry, rounded to nearest whole bill is 99,790,321. The total amount it could carry is:
[tex]A=99,790,321*\$100=\$9,979,032,100[/tex]
What does the term "expand" mean in mathematics?
I am NOT searching for "expanded form" or "distribute".
I think expanding means to remove the parentheses/brackets from a problem.
For example: Say we have the expression: 3 (4 + 5). I think expanding means to multiply 3, by every number in the parentheses. So that means:
(3 * 4) + (3 * 5) = 27.
Another way to think about it is to (if you're on paper) draw a line from 3, to all the numbers inside the parentheses. The line that connects from 3 to 4, is signaling for you to multiply 3 * 4 = 12. And the line from 3 to 5 = 3 * 5 = 15. And add them.
Final answer:
In mathematics, 'expand' refers to writing an expression in an extended form using distribution. This can result in a polynomial or an infinite series, as seen in binomial expansion or exponential arithmetic.
Explanation:
In mathematics, to expand means to increase the length of an expression by distributing multiplication over addition or subtraction. For example, expanding (a + b)(c + d) results in ac + ad + bc + bd. This does not change the value of the expression, but rather writes it in an alternative form that might be more useful for further operations, such as simplification or evaluation. Binomial expansion, specifically, refers to expressing a binomial raised to a power as a series of terms, using the binomial theorem, which can sometimes result in an infinite series or a polynomial of finite length. This expansion is applicable in situations like expanding (x + y)^n or when dealing with power series expansions of standard mathematical functions including exponential arithmetic where numbers are expressed as a product of a digit term and an exponential term such as in the notation 4.57 x 10^3.
dentify the type of data (qualitative/quantitative) and the level of measurement for the native language of survey respondents. Explain your choice. Native language Number of respondents English 759 Spanish 775 French 22 Are the data qualitative or quantitative? A. Quantitative, because descriptive terms are used to measure or classify the data. B. Qualitative, because descriptive terms are used to measure or classify the data. C. Qualitative, because numerical values, found by either measuring or counting, are used to describe the data. D. Quantitative, because numerical values, found by either measuring or counting, are used to describe the data.
Answer:
The correct option is D i.e. Quantitative because numerical values found by either measuring or counting are used to describe the data.
Step-by-step explanation:
As the number of respondents is a numerical value and is identified by counting thus it is a quantitative variable. Also all the other options are incorrect.
A is incorrect because the reason described is not the property of quantitative data.
B is incorrect because the data is not described in descriptive terms.
C is incorrect because the reason described in not a property of qualitative data.
For your own safety, the bank representative says that the following characteristics of PIN numbers are prohibited:________. 1) all four digits identical, 2) sequence of consecutive ascending or descending digits, such as 0123 or 3210, 3) any sequence starting with 19 or 20 (birth years). How many sequences would NOT be permitted?
Answer:
Total = N(1) +N(2)+N(3) = 10+14+200 = 224 sequences
224 sequences would NOT be permitted.
Step-by-step explanation:
1) all four digits identical
Number of possible PIN with all four digits identical N(1) = 1×1×1×10 = 10 possible PIN
2) sequence of consecutive ascending or descending digits N(2) = 14
Only Ascending starting with 0,1,2 = 3
only descending for 7,8,9 = 3
and both for 3,4,5,6 = (2×4)
Total = 3+3+(2×4) = 14
3) any sequence starting with 19 or 20 (birth years
N(3) = 2 × 1×1×10×10 = 2×100 = 200
N(3) = 200
(Note : 100 possible ways for each of 19 or 20s)
Total = N(1) +N(2)+N(3) = 10+14+200 = 224
224 sequences would NOT be permitted.
Calculate descriptive statistics for the variable (Coin) where each of the thirty-five students in the sample flipped a coin 10 times. Round your answers to three decimal places and write the mean and the standard deviation.
Answer:
Mean = 5; Standard Deviation: 1.5811
Step-by-step explanation:
Given Data:
number of times coin flipped = n = 10;
probability of each side of coin = p = 0.5;
Here mean is the product of number of times coin flipped and probability of each
m = n*p =10*0.5 = 5
Standard deviation is obtained by taking square root of product of n,p,q
St. Dev= [tex]\sqrt{npq}[/tex] = [tex]\sqrt{10*0.5*0.5}[/tex] = 1.5811
We have:
Mean = 5 ; Standard Deviation = 1.5811
In a box there are two coins: a standard coin with head and tail and a 2-headed coin. You randomly pick one of the coins, toss it and see a head. What is the probability that the other side of this coin is a head?
Answer:
Probability that the other side of this coin is a head = 0.5
Step-by-step explanation:
Given that there are two coins: a standard coin with head and tail and a 2 - headed coin.
When tossing a randomly chosen coin once the sample space obtained will be :
Head of a standard coin.Tail of a standard coin.Front Head side of 2-headed coin.Back Head side of 2-headed coin.Now since we have to find the probability that the other side of this coin which is tossed is also a head which means probability of selecting a 2-headed coin.
From the above cases there are two outcomes which are in favour from total of four outcomes;
Hence, Probability that the other side of this coin is a head = [tex]\frac{1}{2}[/tex] = 0.5 .
special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin. A. How many elements are there in the sample space? B. Let A be the event that a red card is picked first, followed by landing a tail on the coin toss. P(A) = Present your answer as a decimal number to 1 decimal place. C. Let B be the event that a green or blue is picked, followed by landing a tail on the coin toss. Are the events A and B mutually exclusive? D. Let C be the event that a green or red is picked, followed by landing a tail on the coin toss. Are the events A and C mutually exclusive?
Answer:
Step-by-step explanation:
Given that special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin
A) Sample space will have Green, Head, or Green, Tail .... Red, head, red, tail
No of elements in sample space = no of colours x no of outcomes in coin toss
= 4x2 = 8
B) A= getting (RT)
P(A) = Prob of getting red card and tail on coin
= P (R) *P(T)
=[tex]\frac{4}{20} *\frac{1}{2} \\=\frac{1}{10}[/tex]
C) B be the event that a green or blue is picked, followed by landing a tail on the coin toss
B = getting green card and tail
Getting green card tail is mutally exclusive with red card and tail as there is no common element between green and blue.
D) C= red or green card is picked followed by tail.
Here A and C have a common element as getting red and tail. So not mutually exclusive
Final answer:
The sample space of the card and coin toss experiment consists of 40 elements. The probability of picking a red card followed by a tail coin toss is 0.1. Events A and B, as well as events A and C, are not mutually exclusive.
Explanation:
Probability in a Card and Coin Toss Experiment
When dealing with a special deck of cards and a subsequent coin toss, multiple steps are involved in determining outcomes and probabilities. As for the given student's question:
A). The sample space contains multiple elements based on the card colors and the side of the coin: each of the 20 cards can result in either heads or tails, creating a total of 40 possible outcomes.
B). For event A, where a red card is picked followed by a tails on the coin toss, the probability (P(A)) is calculated by dividing the number of successful outcomes by the number of total possibilities, resulting in P(A) = 4 (red cards) * 1 (tails outcome) / 40 (total outcomes) = 0.1.
C). Event A and event B are not mutually exclusive because event B involves picking either a green or blue card and also getting tails, which does not overlap with the specifics of event A.
D). Similarly, events A and C are not mutually exclusive. Although both involve picking a red card and landing a tail, event C also includes picking a green card, which does not interfere with the occurrence of event A.
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of $1,650. The article also states that 25% of California residents pay more than $1,800.
(a) What is the Z-score that corresponds to the top 25% (or the 75th percentile) of the standard normal distribution? (use the closest value from table B.1)
(b) What is the mean insurance cost? $
What is the cut off for the 75th percentile? $
(c) Identify the standard deviation of insurance premiums in LA.
The Z-score for the 75th percentile is approximately 0.675. The mean insurance cost is $1,650, and the cut off for the 75th percentile is $1,800. The standard deviation of insurance premiums in LA is about $222.22.
Explanation:(a) The Z-score corresponding to the 75th percentile of the standard normal distribution is approximately 0.675. This is determined referring to standard statistical tables or calculators.
(b) The mean insurance cost is given as $1,650.
The cut off for the 75th percentile is $1,800. This is based on the given information that 25% of California residents pay more than $1,800.
(c) To determine the standard deviation, we first subtract the mean from the 75th percentile value ($1,800 - $1,650 = $150), then divide by the Z-score (150 / 0.675 = approximately $222.22). So, the standard deviation for LA auto insurance premiums is about $222.22.
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The Z-score for the 75th percentile of the standard normal distribution is approximately 0.67. The mean insurance cost is $1,650, and the cutoff for the 75th percentile is $1,800. While we can't identify the standard deviation of insurance premiums in LA without additional data, it would be theoretically possible to find it using the Z-score formula.
Explanation:In response to your question, let's take it step by step:
(a): The Z-score that corresponds with the 75th percentile of the standard normal distribution is approximately 0.67. You can find this value by utilizing a look-up table (table B.1).
(b): Given in the question, the mean insurance cost for California residents is $1,650.
The 75th percentile (or cutoff) is $1,800, signifying that 25% of residents pay more than this amount.
(c): Unfortunately, the information provided in the question doesn't offer enough data to figure out the standard deviation for insurance premiums in LA directly. However, based on the details given, you can conclude that for a Z-score of 0.67, the corresponding cost is $1,800. Using the Z-score formula, (X - μ) / σ, where X is the value from the dataset, μ is the mean, and σ is the standard deviation, you could theoretically solve for σ (standard deviation) if all other values are known.
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a 50m long chain hangs vertically from a cunlinder attached to a winch. Assume there is no friction in the system and that the chain has a density of 10kg/m. how much work is required to wind the chain into the cylinder if a 50kg block is attached to the end of the chain?
Answer:
147000 J
Step-by-step explanation:
We are given that
Length of chain=L=50 m
Density of chain=[tex]\rho=10kg/m^3[/tex]
We have to find the work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain.
Work done=[tex]\int_{a}^{b}F(y)dy[/tex]
We have F(y)=[tex]\rho g(50-y)dy[/tex]
a=0 and b=50
[tex]g=9.8m/s^2[/tex]
Using the formula
Work done=[tex]w_1=10\times 9.8\int_{0}^{50}(50-y)dy[/tex]
Where Length of chain is (50-y) has to be lifted.
Work done=[tex]w_1=10\times 9.8[50y-\frac{y^2}{2}]^{50}_{0}[/tex]
By using the formula [tex]\int x^ndx=\frac{x^{n+1}}{n+1}+C[/tex]
Work done=[tex]w_1=10\times 9.8\times (50(50)-\frac{(50)^2}{2})=98\times (2500-1250)=122500 J[/tex]
When the chain is weightless then the work done required to lift the block attached to the 50 m long chain
Again using the formula
Where f(y)=mg
[tex]w_2=\int_{0}^{50}mgdy[/tex]
We have m=50 kg
[tex]w_2=\int_{0}^{50}50\times 9.8 dy=490[y]^{50}_{0}=490\times 50=24500 J[/tex]
The work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain=[tex]w_1+w_2=122500+24500=147000 J[/tex]
For each gym class a school has 10 soccer balls and 6 volleyballs all of the classes share 15 basketballs. The expression 10c+6c+15 represents the total number of balls the school has for c classes what is a simpler form of the expression
Answer:
[tex]16c+15[/tex]
Step-by-step explanation:
we have the expression
[tex]10c+6c+15[/tex]
step 1
We can simplify the expression by combining like terms. That is, the terms with the same variable
[tex](10c+6c)+15[/tex]
[tex]16c+15[/tex]
Answer: 16c+15
Step-by-step explanation:
Step 1
Write down the given expression (10c+6c)+15
Step 2
10+6 = 16c
So, 16c+15
Hope this helps! (✿◡‿◡)
Determine whether the statement is true or false. Justify your answer.
Rolling a number less than 3 on a normal six-sided die has a probability of 1/3. The complement of this event is to roll a number greater than 3, and its probability is 1/2.
Answer:
False
Step-by-step explanation:
The likelihood of an event happening added to its compliment's likelihood must equal 1. The compliment of rolling a number less than 3 on a normal six-sided die is rolling a 3 or more on a normal six-sided die. The probability of rolling 3 or more is 4/6 or 2/3. Adding 1/3 to 2/3 gives us 1 or 100%, thus those events complement each other.
A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in her district.
A) If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
B) If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
Answer:
(A) The minimum sample size required achieve the margin of error of 0.04 is 601.
(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.
Step-by-step explanation:
Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.
(A)
The margin of error, MOE = 0.04.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601[/tex]
Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.
(B)
The margin of error, MOE = 0.02.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401[/tex]
Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.
Use the formula . Find t for r = 33.2 m/h and d = 375.16 m.
[tex]\boxed{t=11.3h}[/tex]
Explanation:In this problem we have the following data:
[tex]r:speed \\ \\ d:distance \\ \\ t:time \\ \\ \\ r=33.2m/h \\ \\ d=375.16m \\ \\ \\ So \ our \ goal \ is \ to \ find \ t[/tex]
Speed, time and distance are related with the following formula:
[tex]r=\frac{d}{t} \\ \\ \\ Solving \ for \ t: \\ \\ t=\frac{d}{r} \\ \\ t=\frac{375.16}{33.2} \\ \\ \boxed{t=11.3h}[/tex]
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Based on a Pitney Bowes survey, assume that 42% of consumers are comfortable having drones deliver their purchases. Suppose we want to find the probability that when five consumers are randomly selected, exactly two of them are comfortable with the drones. What is wrong with using the multiplication rule to find the probability of getting two consumers comfortable with drones followed by three consumers not comfortable, as in this calculation: 10.42210.42210.58210.58210.582 = 0.0344?
Answer:
For this case is wrong use the multiplication for P(X=2):
0.42*0.42*0.58*0.58*0.58 = 0.0344
Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:
[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]
And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=5, p=0.42)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
For this case is wrong use the multiplication for P(X=2):
0.42*0.42*0.58*0.58*0.58 = 0.0344
Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:
[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]
And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.
When running a half marathon (13.1 miles), it took Mark 6 minutes to run from mile marker 1 to mile marker 2, and 20 minutes to run from mile marker 2 to mile marker 4. How long did it take Mark to run from mile marker 1 to mile marker 4? 14/3 minutes 14 3 = 4.666666666666667. What was Mark's average speed as he ran from mile marker 1 to mile marker 4? miles per minute 71 minutes after starting the race, Mark passed mile marker 9. To complete the race in 110 minutes, what must Mark's average speed be as he travels from mile marker 9 to the finish line? miles per minute
Final answer:
It took Mark 26 minutes to run from mile marker 1 to mile marker 4. His average speed was 0.5038 miles per minute for that segment. To complete the race in 110 minutes, Mark's average speed from mile marker 9 to the finish line must be 0.4415 miles per minute.
Explanation:
To find how long it took Mark to run from mile marker 1 to mile marker 4, we need to add the times it took him to run from mile marker 1 to mile marker 2 and from mile marker 2 to mile marker 4. Mark took 6 minutes to run from mile marker 1 to mile marker 2, and 20 minutes to run from mile marker 2 to mile marker 4. So the total time it took him to run from mile marker 1 to mile marker 4 is 6 minutes + 20 minutes = 26 minutes.
To find Mark's average speed as he ran from mile marker 1 to mile marker 4, we need to divide the total distance he ran by the total time it took him. The total distance from mile marker 1 to mile marker 4 is 13.1 miles. So the average speed is 13.1 miles / 26 minutes = 0.5038 miles per minute.
To complete the race in 110 minutes, Mark has 39 minutes left after passing mile marker 9. To find his average speed as he travels from mile marker 9 to the finish line, we need to divide the remaining distance by the remaining time. The remaining distance is 26.22 miles - 9 miles = 17.22 miles. So his average speed is 17.22 miles / 39 minutes = 0.4415 miles per minute.
How many pounds of oranges do the data in the plot line represent?
Answer:
OPTION C: [tex]$ \textbf{37} \frac{\textbf{28}}{\textbf{8}} $[/tex] pounds.
Step-by-step explanation:
From the figure we can see that there are three dots against [tex]$ 3 \frac{7}{8} $[/tex].
That means it becomes [tex]$ 3 \times 3\frac{7}{8} $[/tex].
Note that if there is a mixed fraction of the form [tex]$ a \frac{b}{c} $[/tex] = [tex]$ a + \frac{b}{c} $[/tex].
Therefore, [tex]$ 3 \times 3\frac{7}{8} = 3 \times \bigg(3 + \frac{7}{8} \bigg ) $[/tex] ... (1)
Similarly, against 4 there are 2 dots.
So, it should be [tex]$ 4 \times 2 $[/tex] pounds. ...(2)
3 dots against [tex]$ 4 \frac{1}{8} $[/tex].
So, it becomes [tex]$ 3 \times \bigg(4 + \frac{1}{8} \bigg) $[/tex] ...(3)
Similarly, 2 dots against [tex]$ 4 + \frac{2}{8} $[/tex].
This will become [tex]$ 2 \times \bigg( 2 + \frac{2}{8} \bigg) $[/tex] ...(4)
Now, to calculate the total pound, we simply add (1), (2), (3) & (4).
⇒ [tex]$ 3 \times \bigg(3 + \frac{7}{8} \bigg ) $[/tex] [tex]$ + $[/tex] [tex]$ 4 \times 2 $[/tex] + [tex]$ 3 \times \bigg(4 + \frac{1}{8} \bigg) $[/tex] [tex]$ + $[/tex] [tex]$ 2 \times \bigg( 2 + \frac{2}{8} \bigg) $[/tex]
⇒ [tex]$ 9 + \frac{21}{8} + 8 + 12 + \frac{3}{8} + 8 + \frac{4}{8} $[/tex]
⇒ [tex]$ \bigg ( 9 + 8 + 12 + 8 \bigg) + \bigg( \frac{21 + 3 + 4}{8} \bigg ) $[/tex]
⇒ [tex]$ \textbf{37} \textbf {+} \frac{\textbf{28}}{\textbf{8}} $[/tex] [tex]$ \textbf{=} \hspace{1mm} \textbf{37} \frac{\textbf{28}}{\textbf{8}} $[/tex] which is the required answer.
The revenue of 200 companies is plotted and found to follow a bell curve. The mean is $815.425 million with a standard deviation of $22.148 million. Would it be unusual for a randomly selected company to have a revenue between $747.89 and 818.68 million?1) It is impossible for this value to occur with this distribution of data.2) The value is unusual.
3) We do not have enough information to determine if the value is unusual.
4) The value is not unusual.
5) The value is borderline unusual.
Answer:
option 4
The value is not unusual.
Step-by-step explanation:
If the probability of revenue between $747.89 and $818.68 million is low i.e. less than 5% then we can say that it would be unusual for a randomly selected company to have a revenue between $747.89 and 818.68 million.
We are given that revenue found to follow a bell curve. Also, we are given that mean=815.425 and standard deviation=22.148.
P(747.89<x<818.68)=?
P((747.89-815.425)/22.148<z<(818.68-815.425)/22.148)=P(-3.05<z<0.15)
P(747.89<x<818.68)=0.4989+.0596=0.5585
Thus, the probability for a randomly selected company to have a revenue between $747.89 and 818.68 million is not low and so the value is not unusual.
A woman bought a coat for $99.95 and some gloves for $7.95. If the sales tax was 7%, how much did the purchase cost her? (Round your answer to the nearest cent.)
Answer: $115.5
Step-by-step explanation:
The woman bought a coat for $99.95 and some gloves for $7.95. This means that the total amount of money that the woman would have paid for the coat without tax is
99. 95 + 7.95 = $107.9
If the sales tax was 7%, then the amount paid as sales tax would be
7/100 × 107.9 = 0.07 × 107.9 = $7.553
Therefore, the amount that she would pay to purchase the coat and the gloves is
107.9 + 7.553 = 115.453
= $115.5 rounded up to the nearest cent.
The woman spent
[tex]99.95+7.95=107.9[/tex]
dollars in total. Given the 7% sales tax, it means that she actually paid 107% of this amount. So, the final price was
[tex]107.9\cdot\dfrac{107}{100}=1.079\cdot 107=115.453\approx 115.45[/tex]
dollars.
What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 88 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.) Age range (years) 1-10 11-20 21-30 31 and over Number of individuals 40 15 23 10
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8[/tex]
[tex] s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3[/tex]
[tex]Sd(X) = \sqrt{118.273}=10.9[/tex]
Step-by-step explanation:
Previous concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
For this case we can calculate the properties required with the following table:
Interval Mid point (x) f x*f x^2 *f
_________________________________________
1-10 5.5 40 220 1210
11-20 15.5 15 232.5 3603.75
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
________________________________________
Total 88 1394 32372
We assume that the mid point for the class >31 is 35.5 using the problem information.
For this case the expected value would be given by:
[tex] \bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8[/tex]
The variance owuld be given by this formula"
[tex] s^2 = \frac{n(\sum x^2 f) -(\sum xf)^2}{n(n-1}[/tex]
And if we replace we got:
[tex] s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3[/tex]
The standard deviation would be just the square root of the variance:
[tex]Sd(X) = \sqrt{118.273}=10.9[/tex]
To compute mean age, sample variance, and standard deviation, establish class midpoints, then follow steps of calculating mean, variance, and standard deviation methodologies. Utilize relevant class midpoints, frequencies and population size.
Explanation:Given the prehistoric Native American family group of 88 members, we can compute the mean age, sample variance, and sample standard deviation using the group's age ranges and sizes.
First, determine the midpoint for each age range: 5.5 (for 1-10), 15.5 (for 11-20), 25.5 (for 21-30), and 35.5 (for 31 and over).Then, multiply these midpoints by the number of individuals in each group and sum these products to calculate the total age.The mean age is obtained by dividing the total age by the total number of individuals (88).To find the sample variance, first compute the square of the difference between each class midpoint and the mean for all data points. Multiply each squared difference by the corresponding class frequency, add these products together, then divide by (n - 1).The sample standard deviation is the square root of the sample variance.Learn more about Statistics here:https://brainly.com/question/31538429
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Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the Poisson distribution. Provide answers to the following to 3 decimal places. What is the probability that exactly two customers arrive in a minute? Find the probability that more than three customers arrive in a two-minute period. What is the probability that at least seven customers arrive in three minutes, given that exactly two arrive in the first minute?
Answer:
a) P(2)=0.270
b) P(X>3)=0.605
c) P=0.410
Step-by-step explanation:
We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:
[tex]\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}[/tex]
a) In this case: [tex]\lambda=2.1[/tex]
[tex]P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270[/tex]
Therefore, the probability is P(2)=0.270.
b) In this case: [tex]\lambda=2\cdot 2.1=4.2[/tex]
[tex]P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605[/tex]
Therefore, the probability is P(X>3)=0.605.
c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.
In this case: [tex]\lambda=2\cdot 2.1=4.2[/tex]
[tex]P(X\geq 5)=1-P(X<5)\\\\P(X\geq 5)=1-P(X\leq 4)\\\\P(X\geq 5)=1-\sum_{x=0}^4 \frac{4.2^x\cdot e^{-4.2}}{x!}\\\\P(X\geq 5)=1-0.590\\\\P(X\geq 5)=0.410[/tex]
Therefore, the probability is P=0.410.
Suppose X indicates the number of customers that enter a grocery store within one minute.
[tex]\to X \sim \text{Poisson}(2.1)[/tex]
X's probability mass function:
[tex]\to P(X=x)=\frac{e^{-2.1} \times 2.1^{x}}{x!} ,x =0,1,2,..[/tex]
For question 1:
The likelihood of exactly two consumers arriving in a minute:
[tex]P(X=2)=\frac{e^{-2.1} \times 2.1^{2}}{2!}= 0.270016 \approx 0.270[/tex]
For question 2:
Suppose Y represents the average group of consumers who enter a grocery shop in a two-minute period.
[tex]Y \sim \text{Poisson}(2.1\times 2) \ or\ Y \sim \text{Poisson}(4.2)[/tex]
Y's probability mass function:
[tex]P(Y = y) = \frac{e^{-4.2} \times 4.2^{y}}{y!}, y =0,1,2..[/tex]
This is possible that more than three clients will arrive within a two-minute timeframe.
[tex]= P(Y > 3) \\\\= 1 - P(Y \leq 3)\\\\=1- \Sigma^{3}_{y=0} \frac{e^{-4.2} \times 4.2^y}{y!}\\\\=1-0.395403\\\\= 0.604597\approx \ 0.605[/tex]
For question 3:
Given that exactly two customers arrive in the first minute, the likelihood of at least seven clients arriving in three minutes is
[tex]= \text{P( at least 5 customers arrive in two minutes)} \\\\= P(Y \geq 5)\\\\= 1 - P(Y<5)\\\\=1-P(Y \leq 4)\\\\= 1 - 0.589827\\\\= 0.410173 \approx \ 0.410[/tex]
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Taylor, Moore, and Jenkins are candidates for public office. It is estimated that Moore and Jenkins have about the same probability of winning, and Taylor is believed to be twice as likely to win as either of the others. Find the probability of each candidate winning the election.
Answer:
Taylor = 50%
Moore = 25%
Jenkins = 25%
Step-by-step explanation:
Assuming there are no other candidates and that someone has to win the election, the probabilities of Taylor, Moore, and Jenkins winning the election must add up to 1 or 100%.
[tex]T+M+J = 1\\[/tex]
Since Moore and Jenkins have about the same probability of winning, and Taylor is believed to be twice as likely to win as either of the others:
[tex]M=J\\T=2J\\J+J+2J=1\\J=0.25\\M=J=0.25\\T=2*0.25=0.5[/tex]
Taylor has a probability of 50% of winning the election.
Moore has a probability of 25% of winning the election.
Jenkins has a probability of 25% of winning the election.
According to the National Household Survey on Drug Use and Health, when asked in 2012, 41% of those aged 18 to 24 years used cigarettes in the past year, 9% used smokeless tobacco, 36.3% used illicit drugs, and 10.4% used pain relievers or sedatives. Explain why it is not correct to display these data in a pie chart.
a. The types of illicit drugs are not given.
b. There could be roundoff error.
c. The three groups do not add up to 100%.
d. There have to be more than three categorical variables.
e. There could be overlap between the groups.
Final answer:
The correct answer is c. The three groups do not add up to 100%. A pie chart is used to display the parts of a whole, where each category represents a proportion of the total. In this case, the categories of cigarette use, smokeless tobacco use, illicit drug use, and pain reliever/sedative use do not add up to 100% when combined. As a result, a pie chart would not accurately represent the data.
Explanation:
The correct answer is c. The three groups do not add up to 100%.
A pie chart is used to display the parts of a whole, where each category represents a proportion of the total. In this case, the categories of cigarette use, smokeless tobacco use, illicit drug use, and pain reliever/sedative use do not add up to 100% when combined. As a result, a pie chart would not accurately represent the data.
what is the value of 7 to the 4th power
Answer:
The value is 2401
Step-by-step explanation:
First we need to understand what's means 7 to the 4th power, or 7^4.
7 is the power base, the power base is the number that we are going to repeat the number of times the exponent says.
4 is the exponent, which will tell us how many times to multiply 7.
So , then we would have
1) 7 x 7 x 7 x 7
2) 49 x 7 x 7
3) 343 x 7
4) finally 2401