A government bureau keeps track of the number of adoptions in each region. The accompanying histograms show the distribution of adoptions and the population of each region. ​a) What do the histograms say about the​ distributions? ​b) Why do the histograms look​ similar? ​c) What might be a better way to express the number of​ adoptions?

Answer:

0.94

Step-by-step explanation:

System will function if both components function, so,

P(system function)=P(A∩B)=?

P(A∩B)=P(A)+P(B)-P(A∪B)

We are given that P(A)=0.98, P(B)=0.95 and P(A or B)=P(A∪B)=0.99.

P(A∩B)=0.98+0.95-0.99=1.93-0.99=0.94

P(system function)=P(A∩B)=0.94.

Thus, the probability that the system functions is 0.94 or 94%.

The probability that the system functions, requiring both A and B to function, is calculated using the formula for the probability of either event occurring. By rearranging and substituting the given probabilities, we find that the probability the system functions is 0.94.

To determine the probability that the system functions, we need to find the joint probability that both A and B function, denoted as P(A AND B). Given that the probability A functions is 0.98, and B functions is 0.95, we use the given that the probability either A or B functions (which includes the case where both function) is 0.99.

We start with the formula for the probability that either A or B functions, which is:

P(A OR B) = P(A) + P(B) − P(A AND B)

.

We can rearrange this to solve for P(A AND B):

P(A AND B) = P(A) + P(B) − P(A OR B)

.

Substituting the given probabilities, we get:

P(A AND B) = 0.98 + 0.95 − 0.99 = 0.94

.

Therefore, the probability that the system functions, which requires both A and B to function, is 0.94.

Answer:

a) Option D is correct for this question.

That is, A or B is the event that the person is overweight or obese.

But for other questions whose sets aren't disjoint, event A or B usually means all the elements that are in either A or B or both sets.

b) Option C.

P(A or B) = 0.71

c) Option A

P(C) = 1 - P(A or B) = 0.29

Step-by-step explanation:

b) Event B specifically rules out obesity, meaning, set A and set B have no elements in common.

In a normal probability question, event A or B usually means all the elements that are in either A or B and elements that are in the two sets.

But for this question, since, it has been made clear that there are no common elements in the two sets, event A or B is event that the person is overweight or obese. Option D.

b) For disjoint sets, P(A or B) = P(A) + P(B) = 0.38 + 0.33 = 0.71. Option C.

c) P(C) is the set of all elements that are not in either A or B.

P(C) = 1 - P(A or B) = 1 - 0.71 = 0.29. Option A.

Answer:

(6.8,8)

Step-by-step explanation:

mean=7.4

standard deviation=0.2

we have to find the range in which at least 88.9% of the data will reside

1-1/k²=0.889

1/k²=1-0.889

1/k²=0.111

k²=1/0.111=9.009

k=3.002

so, k=3

The range of values can be computed as mean±k(standard deviation).

Thus, the range in which at least 88.9% of the data will reside is

(mean-k(standard deviation), mean+ k(standard deviation))

(7.4-3(0.2),7.4+3(0.2))

(7.4-0.6,7.4+0.6)

(6.8,8)

Thus, the range in which at least 88.9% of the data will reside is (6.8,8).

The median of a set of bids can be found by arranging them in numerical order and selecting the middle value.

The median is the middle value of a set of data arranged in numerical order. In this case, we have a set of six bids: 2.2, 1.3, 1.9, 1.2, 2.4, and x (blurred number). To find the median, we first need to arrange the bids in numerical order:

Since there are six bids, the middle value will be the fourth number in the ordered list. Therefore, the median is 2.2.

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Answer:

Answer: 40 sq. in.

Step-by-step explanation:

First we gotta find the area of triangle part of the shape, we can see the left side of the shape is 5 in. , so we subtract the 3 from 5 which gives 2 the height of the triangle.

Now, we find the length for the base of the triangle, the top part of the shape is 7 in. , so we subtract 7 from 12 which gives us 5 in. as the base

Now, we find the area of the triangle:

A = [tex]\frac{1}{2}[/tex]b × h

A = ([tex]\frac{1}{2}[/tex] × 5 in,)  × 2 in

A = 5 sq. in.

Now we find the area for the rectangle:

A = b × h

A = 5 x 7

A = 35 sq. in

Finally, we add the areas together

35 sq. in. + 5 sq. in. = 40 sq. in.

We get our answer 40 sq. in,

Answer:

19%

Step-by-step explanation:

We assume that the class has to have a president. So the chances of none of the candidates winning is 0%. In turn we expect that one of the candidates will win the elections, therefore the chances of someone winning is 100%. Therefore the chances of the  third candidate winning can be calculated by removing the chances of the other two candidates winning

P(third candidate winning) = 100% - (37%+44%) = 19%

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

To solve this problem, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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 = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85[/tex]

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{177.6 - 185}{3.85}[/tex]

[tex]Z = -1.92[/tex]

[tex]Z = -1.92[/tex] has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of [tex]Z = 1.92[/tex]

[tex]Z = 1.92[/tex] has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Answer:

for sample = xbar

population = μ

Step-by-step explanation:

The arithmetic mean for sample can be represented by xbar and it can be calculated as

xbar=∑xi/n

Where xi represents data values and n represents number of data values in a sample.

The arithmetic mean for population can be represented by μ and it can be calculated as

μ=∑xi/N

Where xi represents data values and N represents number of data values in a population.

The symbol for the arithmetic mean when it's a sample statistic is 'x', and 'μ' when it's a population parameter. Sample mean (x) refers to the mean of a sample group, while population mean (μ) refers to the mean of an entire population.

The symbol that is used for the arithmetic mean when it is a sample statistic is 'x' (pronounced as x-bar). This is used to denote the mean of a sample. On the other side, the symbol that is used when the arithmetic mean is a population parameter is 'μ' (pronounced as mu). This symbol is representative of the mean of an entire population.

Take for example a group of 50 students in a class who took a test. If you wanted to find the sample mean (x) of their scores, you'd add all their scores and divide by the total number of students (which is 50 in this case). However, if you were to calculate the population mean (μ) of the scores of all students in all high school classes in the nation, you'd add their scores and divide by the total number.

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Answer:

a) there is s such that r>s and s is positive

b) For any r>0 , there exists s>0 such that s<r

Step-by-step explanation:

a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.

b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.

The statement with variables filled in is: (a) Given any positive real number r, there is s such that s is a positive real number and s is less than r. (b) For any positive real number r, there exists s such that s < r.

The statement can be rewritten with variables as follows:

(a) Given any positive real number r, there is s such that s is a positive real number and s is less than r.

(b) For any positive real number r, there exists s such that s < r.

The concept here is that for any positive real number, you can always find another positive real number that is less than the given number. This is fundamental in understanding the density of real numbers on the number line, where between any two distinct real numbers, there are infinitely many other real numbers.

Answer: 56 outcomes

Step-by-step explanation:

Tossing a die 3 times and not having the order recorded, we have the following outcome.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,4] [1,4,5] [1,4,6]

[1,5,5] [1,5,6]

[1,6,6]

[2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,4] [2,4,5] [2,4,6]

[2,5,5] [2,5,6]

[2,6,6]

[3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,4] [3,4,5] [3,4,6]

[3,5,5] [3,5,6]

[3,6,6]

[4,4,4] [4,4,5] [4,4,6]

[4,5,5] [4,5,6]

[4,6,6]

[5,5,5] [5,5,6]

[5,6,6]

[6,6,6]

Hence, By direct counting, the number of possible outcome is 56 outcomes.

Answer:

a) the length of the wire for the circle = [tex](\frac{60\pi }{\pi+4}) in[/tex]

b)the length of the wire for the square = [tex](\frac{240}{\pi+4}) in[/tex]

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=[tex]\frac{y}{4}[/tex]

Area of the square which is L² can now be said to be;

[tex]A_S=(\frac{y}{4})^2 = \frac{y^2}{16}[/tex]

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

[tex]r = \frac{60-y}{2\pi }[/tex]

Area of the circle which is πr² can now be;

[tex]A_C= \pi (\frac{60-y}{2\pi } )^2[/tex]

     [tex]=( \frac{60-y}{4\pi } )^2[/tex]

Total Area (A);

A = [tex]A_S+A_C[/tex]

   = [tex]\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2[/tex]

For the smallest possible area; [tex]\frac{dA}{dy}=0[/tex]

∴ [tex]\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0[/tex]

If we divide through with (2) and each entity move to the opposite side; we have:

[tex]\frac{y}{18}=\frac{(60-y)}{2\pi}[/tex]

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

[tex]y= \frac{240}{\pi+4}[/tex]

∴ since [tex]y= \frac{240}{\pi+4}[/tex], we can determine for the length of the circle ;

60-y can now be;

= [tex]60-\frac{240}{\pi+4}[/tex]

= [tex]\frac{(\pi+4)*60-240}{\pi+40}[/tex]

= [tex]\frac{60\pi+240-240}{\pi+4}[/tex]

= [tex](\frac{60\pi}{\pi+4})in[/tex]

also, the length of wire for the square  (y) ; [tex]y= (\frac{240}{\pi+4})in[/tex]

The smallest possible area (A) = [tex]\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})[/tex]

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

Answer:

S= {(A,B,C) , (A,C,B) , (B,C,A), (B,A,C), (C,A,B), (C,B,A)}

And we have 6 possible outcomes for this case.

Step-by-step explanation:

By definition the sample space of an experiment "is the set of all possible outcomes or results of that experiment".

For the case described here: "A taste tester ranks three varieties of yogurt, A, B, and C, according to preference.".

Assuming that we have three varieties of yogurt {A,B,C}

We denote the event (A,B,C) like this: A is the first, B the second and C the third in the rank. And for example the event (C,B,A) means that C is on the 1th position of the rank, B on the 2nd position and A on the 3th position.  

The sampling space denoted by S and is given by:

S= {(A,B,C) , (A,C,B) , (B,C,A), (B,A,C), (C,A,B), (C,B,A)}

And we have 6 possible outcomes for this case.

The sample space for ranking three yogurt varieties (A, B, C) consists of 6 permutations: (A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), and (C, B, A).

The problem involves finding all possible ways a taste tester can rank three varieties of yogurt: A, B, and C. In this context, a sample space represents all possible outcomes of the experiment.

Since the taste tester is ranking the yogurt varieties, we are looking for all permutations of the set {A, B, C}.

A permutation is an arrangement of all the members of a set in a specific order. For three items, the total number of permutations is given by 3! (3 factorial), which equals 6.

Therefore, the sample space S, containing all possible rankings, is:

Answer:buy here both

Step-by-step explanation:

Answer:

y = -3000x + 25,635

Step-by-step explanation:

well if the inital value of the car is $25,635 this means that when

x = 0   y = 25,635

(0 , 25635)

this will be our first point

Now if you tell us that in 1 year depreciate in value $3,000

this means thaw when

x = 1  y = 25,635 - 3,000

x = 1 y = 22,635

(1, 22635)

Now that we have 2 points we can have the equation

First we take the slope as follows

m = (y2 - y1) / (x2 - x1)

m = (22,635 - 25,635) / (1 - 0)

m = -3000 / 1

m = -3000

after calculating the slope we have to replace it in the following formula

(y - y1) = m (x - x1)

y - 25,635 = -3000 ( x - 0)

y - 25,635 = -3000x

y = -3000x + 25,635

Finally we replace the value of y by 10000

10,000 = -3000x + 25,635

10,000 - 25,635 = -3000x

-15,635 = -3000x

-15,635/-3000 = x

5.21167 = x       years

These are the years it would take for the value to be 10,000

to know the days we simply multiply by 365

5.21167 * 365 = 1902.26      days

Answer:

5 classes.

Step-by-step explanation:

You can use the [tex]2^k[/tex] rule to determine the number of classes for a frequency distribution.

The [tex]2^k[/tex] rule says that [tex]2^k\geq n[/tex] where

[tex]k[/tex] is the number of classes

[tex]n[/tex] is the number of the data points

We know that the number of data points is [tex]n[/tex] = 20.

Next, we start searching for [tex]k[/tex] so that we can get a number 2 to the [tex]k[/tex] that is larger that the number of data points.

[tex]2^4=16\\2^5=32[/tex]

This suggests that you should use 5 classes.

Final answer:

For this data, it is recommended to use 5 classes.

Explanation:

To organize the data into a frequency distribution, you need to determine the number of classes.

The recommended number of classes can vary depending on the data.

One popular rule is to use the square root of the number of data points to determine the number of classes.

In this case, there are 20 data points, so the square root is approximately 4.47.

Since you can't have a fraction of a class, you can round up to the nearest whole number, which gives you a recommendation of 5 classes.

Answer:

The probability that there was no messages received during a 2-hour period is 0.3679.

Step-by-step explanation:

Let the random variable X = time between the arrival of e-mail messages.

The random variable [tex]X\sim Exp(\lambda)[/tex]

The probability distribution function of exponential distribution is:

[tex]f(x)=\left \{ {{\lambda e^{-\lambda x};\ x>0} \atop {0};\ otherwise} \right.[/tex]

The mean of the distribution is, Mean = 2.

The value of λ is:

[tex]\lambda=\frac{1}{Mean} =\frac{1}{2}=0.50[/tex]

Compute the probability that there was no messages received during a 2-hour period as follows:

[tex]P(X>2)=1-P(X\leq 2)\\=1-\int\limits^{2}_{0} {\lambda e^{-\lambda x}} \, dx \\=1-\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{2}_{0}\\=1-[1-e^{-\frac{x}{2} }]^{2}_{0}\\=1-[1-e^{-\frac{2}{2}}]\\=e^{-1}\\=0.3679[/tex]

Thus, the probability that there was no messages received during a 2-hour period is 0.3679.

Answer:

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

    2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x [tex]10^{-8}[/tex] .  

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

                  Total containers = 445

                   Defective ones   = 3

           Non - Defective ones = 442 { Acceptable ones}

(a) Probability that the second one selected is defective given that the first one was defective is given by;

  Since we had selected one defective so for selecting second the available

  containers are 444 and available defective ones are 2 because once

   chosen they are not replaced.

Hence, Probability that the second one selected is defective given that the first one was defective = [tex]\frac{2}{444}[/tex] = 0.00450

(b) Probability that both are defective = P(first being defective) +

                                                                     P(Second being defective)

                 = [tex]\frac{3}{445} + \frac{2}{444}[/tex] = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) +  P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

 So, Required Probability = [tex]\frac{442}{445}*\frac{441}{444}[/tex] = 0.986

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = [tex]\frac{1}{443}[/tex] = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.

Hence, Required probability = [tex]\frac{2}{443}[/tex] = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

                              P(Second being defective) +  P(Third being defective)

        = [tex]\frac{3}{445}* \frac{2}{444} * \frac{1}{443}[/tex] = 6.855 x [tex]10^{-8}[/tex] .                

Final answer:

The probability values are found by considering the number of defective and non-defective containers remaining at each step of selection.

Explanation:

The probability that the second one selected is defective given that the first one was defective can be found by looking at the remaining pool of containers. Initially, there are 3 defective containers out of 445.

After picking one, there are 2 defective containers left out of 444. The probability is therefore calculated as 2/444 = 0.00450 (rounded to five decimal places).

The probability that both containers are defective is calculated by multiplying the probability of selecting a defective container on the first draw with the probability of selecting a defective container on the second draw: (3/445) * (2/444) = 0.00003 (rounded to seven decimal places).

The probability that both containers are acceptable is the probability of not selecting a defective container twice: (442/445) * (441/444) = 0.98876.

If the first and second containers selected are defective, the probability that the third one selected is defective comes from the remaining one defective container out of 443: 1/443 = 0.00226.

If the first container is defective and the second one is not, the probability that the third one selected is defective is 2/443 = 0.00451 (rounded to five decimal places).

The probability that all three containers selected are defective is the product of the probabilities of selecting a defective container each time without replacement: (3/445) * (2/444) * (1/443) = 0.00002.

Answer:

$129 000/yr

Step-by-step explanation:

Weighted average is used to answer this question.

total employees are 20

10 employees make 80 000

total earning for 10 employees = 80 000 * 10 = 800 000 (multiplying)

6 employees make 150 000

total earning for 6 employees= 150 000 * 6 =900 000 (multiplying)

4 employees make 220 000

total earning for 4 employees = 220 000 * 4 = 880 000 (multiplying)

To calculate weighted average all the totals are added and then divide by total number of employees.

weighted average = (800 000 + 900 000 + 880 000)/20

weighted average = 2580000/20

Weighted average = 129 000

Answer:

4.5 and 8.5 hours.

Step-by-step explanation:

The empirical 95% rule states that, in a normal distribution, 95% of the data falls within two standard deviations bellow or above the mean. If the mean is 6.5 hours and the standard deviation is 1 hour, the interval is:

[tex]6.5-2*1\leq x \leq 6.5+2*1\\4.5\leq x \leq 8.5[/tex]

The range that contains the middle 95 % of hours slept per week night by college students is 4.5 to 8.5 hours.

Using the concept of Normal Distribution, and knowing that 95% of data falls within two standard deviations, we find that hours of sleep among students range between 4.5 and 8.5 hours.

This question involves the concept of a Normal distribution in statistics, commonly used in analyzing patterns of data spread. In this scenario, we know the sleep hours are normally distributed with a mean (average) of 6.5 hours and a standard deviation of 1 hour. The middle 95% of hours slept per week night by college students means we are looking for the range of sleep hours that falls within two standard deviations of the mean on a normal distribution curve.

Because the standard deviation is 1 hour, two standard deviations would be 2 hours. Thus, we subtract and add two standard deviations from the mean to find the range. That is, 6.5 - 2 = 4.5 hours (lower bound) and 6.5 + 2 = 8.5 hours (upper bound).

So, the middle 95% of sleep hours per week night by college students falls within the range of 4.5 to 8.5 hours.

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Answer:

V=15.44

Step-by-step explanation:

We have a formula

V=\int^{π/3}_{-π/3} A(x) dx ,

where A(x) calculate as  cross sectional.

We have:

Inner radius: 5 + sec(x) - 5= sec(x)  

Outer radius: 7 - 5=2, we get

A(x)=π 2²- π· sec²(x)

A(x)=π(4-sec²(x))

Therefore, we calculate the volume V, and we get

V=\int^{π/3}_{-π/3} A(x) dx

V=\int^{π/3}_{-π/3} π(4-sec²(x)) dx

V=[ π(4x-tan(x)]^{π/3}_{-π/3}

V=π·(8π/3-2√3)

V=15.44

We use a site geogebra.org to plot the graph.

The volume of the solid obtained by rotating the region bounded by y = 5 + sec(x), -π/3 ≤ x ≤ π/3, y = 7 about y = 5 is calculated using the disc method formula in calculus, resulting in a volume of 8π²/3 cubic units.

The region that we need to rotate around the line y=5 is bounded by the curves y=7 and y=5+sec(x) over the interval -π/3 ≤ x ≤ π/3. The resulting solid is a disc shaped object. We can find the volume of this solid using the disc method formula in calculus:

V = π ∫ [R(x)]² dx

, where R(x) = radius function. The radius function is the absolute difference between y=7 (the upper curve) and y=5 (the line of rotation), which equals 2. Therefore, the integral becomes:

V = π ∫ from -π/3 to π/3 [2]² dx

, which simplifies to

V = 4π [x] from -π/3 to π/3

. Finally, evaluate this expression by subtracting the lower limit from the upper limit, giving

V = 4π(2π/3) = 8π²/3

cubic units.

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Answer:

[tex]y(t) = \$1900*1.0375^t[/tex]

Step-by-step explanation:

If y is the future value of the $1900 investment, in dollars, after t years at a rate of 3.75% per year, compounded annually. The exponential function that describes the relationship between the variables y and t is:

[tex]y(t) = \$1900*(1.0375)^t[/tex]

This relationship means that for every year t, the amount y will increase by a factor of 1.0375.

To represent the value of an investment of $1900 at a 3.75% annual interest rate compounded annually, the exponential function y = 1900(1 + 0.0375)^t is used, where t is the number of years.

The question asks for an exponential function that shows the relationship between the value of an investment, y, and time, t, when the principal is $1900 and the interest rate is 3.75% compounded annually. The standard formula for an investment compounded annually is y = P(1 + r)t, where P is principal, r is the annual interest rate as a decimal, and t is the number of years.

For this specific case, the function would be y = 1900(1 + 0.0375)t. This represents the value of the investment after t years, given the initial investment and interest rate provided.

Answer: Check the attached

Step-by-step explanation:

Total distance traveled from the starting point = 487 mm (or) 0.487 m

Solution:

Given data:

Length of each block = 135 mm

Distance traveled towards North = three blocks

                                                      = 135 × 3

                                                      = 405

Distance traveled towards North = 405 mm

Distance traveled towards West = two blocks

                                                      = 135 × 2

                                                      = 270

Distance traveled towards West = 270 mm

Total distance traveled

               [tex]=\sqrt{\text{Distance traveled in North}^2+\text{Distance traveled in West}^2}[/tex]

               [tex]=\sqrt{405^2+270^2}[/tex]

               [tex]=\sqrt{236925}[/tex]

               [tex]=135\sqrt{13}[/tex]

               = 487 mm (approximately)

Total distance traveled from the starting point = 487 mm

Let us convert mm to m.

1 m = 1000 mm

487 mm = 487 ÷ 1000 = 0.487 m

Total distance traveled from the starting point = 0.487 m

Answer:

C. Data Mining

Step-by-step explanation:

As data mining is a technique which is used predict and to find patterns or relationships among elements of the data in a large database. It also facilitate the enterprise to predict the future trends.

a. The average velocity for the time period beginning at t = 2 and lasting 0.5 seconds is -36 ft/s.

b. The average velocity for the time period beginning at t = 2 and lasting 0.1 seconds is -29.6 ft/s.

c. The average velocity for the time period beginning at t = 2 and lasting 0.05 seconds is -40.8 ft/s.

d. The average velocity for the time period beginning at t = 2 and lasting 0.01 seconds is 3.84 ft/s.

e. The estimated instantaneous velocity at t = 2 is -32 ft/s.

(a) To find the average velocity over a time period, we use the formula for average velocity:

[tex]\[ \text{Average Velocity} = \frac{\text{Change in height}}{\text{Change in time}} \][/tex]

Given [tex]\( y = 36t - 16t^2 \)[/tex], we'll find the heights at [tex]\( t = 2 \)[/tex] and [tex]\( t = 2.5 \)[/tex], then calculate the difference.

At [tex]\( t = 2 \), \( y = 36(2) - 16(2)^2 = 72 - 64 = 8 \)[/tex] ft.

At [tex]\( t = 2.5 \), \( y = 36(2.5) - 16(2.5)^2 = 90 - 100 = -10 \)[/tex] ft.

Change in height = [tex]\( -10 - 8 = -18 \)[/tex] ft.

Change in time = [tex]\( 2.5 - 2 = 0.5 \)[/tex] sec.

Average velocity = [tex]\( \frac{-18}{0.5} = -36 \)[/tex] ft/s.

(b) Following the same procedure as (a), at [tex]\( t = 2.1 \), \( y = 36(2.1) - 16(2.1)^2 = 75.6 - 70.56 = 5.04 \)[/tex] ft.

Change in height = [tex]\( 5.04 - 8 = -2.96 \)[/tex] ft.

Change in time = [tex]\( 2.1 - 2 = 0.1 \)[/tex] sec.

Average velocity = [tex]\( \frac{-2.96}{0.1} = -29.6 \)[/tex] ft/s.

(c) Continuing the process, at [tex]\( t = 2.05 \), \( y = 36(2.05) - 16(2.05)^2 = 73.8 - 67.84 = 5.96 \)[/tex] ft.

Change in height = [tex]\( 5.96 - 8 = -2.04 \)[/tex] ft.

Change in time = [tex]\( 2.05 - 2 = 0.05 \)[/tex] sec.

Average velocity = [tex]\( \frac{-2.04}{0.05} = -40.8 \)[/tex] ft/s.

(d) Applying the method to [tex]\( t = 2.01 \), \( y = 36(2.01) - 16(2.01)^2 = 72.36 - 64.3216 = 8.0384 \)[/tex] ft.

Change in height = [tex]\( 8.0384 - 8 = 0.0384 \)[/tex] ft.

Change in time = [tex]\( 2.01 - 2 = 0.01 \)[/tex] sec.

Average velocity = [tex]\( \frac{0.0384}{0.01} = 3.84 \)[/tex] ft/s.

(e) To estimate the instantaneous velocity at t = 2, we find the derivative of [tex]\( y = 36t - 16t^2 \)[/tex] with respect to [tex]\( t \)[/tex], which is [tex]\( v(t) = 36 - 32t \)[/tex]. Plugging in [tex]\( t = 2 \)[/tex],

we get [tex]\( v(2) = 36 - 32(2) = -32 \)[/tex] ft/s.

Answer:

a) [tex] P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667[/tex]

Represent the probability that the event B occurs given that the event A occurs first

b) [tex] P(B'|A) = \frac{0.15}{0.45}=0.333[/tex]

Represent the probability that the event B no occurs given that the event A occurs first

c) [tex] P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857[/tex]

Represent the probability that the event A occurs given that the event B occurs first

d) [tex] P(A'|B) = \frac{0.05}{0.35}=0.143[/tex]

Represent the probability that the event A no occurs given that the event B occurs first

Step-by-step explanation:

For this case we have the following probabilities given for the events defined A and B

[tex] P(A) = 0.45, P(B) = 0.35, P(A \cap B) =0.30[/tex]

For this case we can begin finding the probability for the complements:

[tex] P(B') =1-P(B) = 1-0.35= 0.65[/tex]

[tex] P(A') =1-P(A) = 1-0.45= 0.55[/tex]

For this case we are interested on the following probabilities:

Part a

[tex] P(B|A)[/tex]

For this case we can use the Bayes theorem and we can find this probability like this:

[tex] P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667[/tex]

Represent the probability that the event B occurs given that the event A occurs first

Part b

[tex] P(B'|A) = \frac{P(B' \cap A)}{P(A}[/tex]

And for this case we can find [tex] P(B' \cap A) =P(A) -P(A\cap B)= 0.45-0.3=0.15[/tex]

And if we replace we got:

[tex] P(B'|A) = \frac{0.15}{0.45}=0.333[/tex]

Represent the probability that the event B no occurs given that the event A occurs first

Part c

[tex] P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857[/tex]

Represent the probability that the event A occurs given that the event B occurs first

Part d

[tex] P(A'|B) = \frac{P(A' \cap B)}{P(B}[/tex]

And for this case we can find [tex] P(A' \cap B) =P(B) -P(A\cap B)= 0.35-0.3=0.05[/tex]

And if we replace we got:

[tex] P(A'|B) = \frac{0.05}{0.35}=0.143[/tex]

Represent the probability that the event A no occurs given that the event B occurs first

Conditional probabilities are calculated as the ratio of the intersection of two events to the probability of the given event. P(B|A) is approximately 0.6667, P(B'|A) is approximately 0.3333, P(A|B) is approximately 0.8571, and P(A'|B) is approximately 0.1429.

When considering the probability of randomly selecting a student at a university who has a Visa (Event A) or a MasterCard (Event B), we can use the given probabilities to calculate the following:

P(B|A) is the probability that a selected individual has a MasterCard given they have a Visa. It is calculated as P(A [tex]\cap[/tex] B) / P(A). Given P([tex]A \cap B[/tex]) = 0.30 and P(A) = 0.45, P(B|A) = 0.30 / 0.45 which rounds to 0.6667.

P(B'|A) is the probability that a selected individual does not have a MasterCard given they have a Visa. It is 1 - P(B|A), which is 1 - 0.6667, rounding to 0.3333.

P(A|B) is the probability that a selected individual has a Visa given they have a MasterCard. It is calculated as P(A \\cap B) / P(B). Given P(A [tex]\cap[/tex] B) = 0.30 and P(B) = 0.35, P(A|B) = 0.30 / 0.35 which rounds to 0.8571.

P(A'|B) is the probability that a selected individual does not have a Visa given they have a MasterCard. It is 1 - P(A|B), which is 1 - 0.8571, rounding to 0.1429.

Answer:

a. Mean doesn't change.

b. Median doesn't change.

c. Mode can change.

Step-by-step explanation:

Let us assume the data set with 10 observations

{2,6,4,3,2,6,4,9,4,7}.

Arranging data set in ascending order

{2,2,3,4,4,4,6,6,7,9}

mean=2+2+3+4+4+4+6+6+7+9/10=4.7

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Mode

The most repeated value is 4. So, mode is 4 for assumed data.

Increasing highest value by 10 and decreasing lowest value by 10

{-8,2,3,4,4,4,6,6,7,19}

a.

mean=-8+2+3+4+4+4+6+6+7+19/10=4.7

Mean doesn't change

b.

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Median doesn't change

c.

Most occurring value is still 4. But mode can change if the value the highest value becomes most concurring value.

A. The mean may or may not change, depending on the original data values. If the data set has an odd number of values and the highest and lowest values are not the same, then the mean will change.

B. The median will not change because it is the middle value of the data set, and adding or subtracting a constant value to all data points does not affect the relative order of the values. The position of the median will remain the same

C. The mode may or may not change. If the mode was the highest or the lowest value in the original data set and its frequency did not change after the adjustments, the mode will remain the same.

A.However, if the data set has an even number of values or the highest and lowest values are the same, then the mean will remain unchanged. For example:

Original data set: 5, 8, 10, 12, 15

After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5

Mean before: (5+8+10+12+15)/5 = 10

Mean after: (25+8+10+12+(-5))/5 = 10

B. For example: Original data set: 5, 8, 10, 12, 15

After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5

Median before: 10 (middle value of the ordered data set)

Median after: 10 (still the middle value of the ordered data set)

C. However, if the highest or lowest value was not the mode in the original data set and its frequency becomes the highest after the adjustments, the mode will change. For example:

Original data set: 5, 8, 10, 12, 15

After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5

Mode before: No mode (all values are unique)

Mode after: 10 (frequency of 10 is now higher than other values)

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Final answer:

The probabilities in a special deck of 16 cards for drawing a card that is a two or a four are as follows: (a) 1/2, (b) 2/3, (c) 1/2, and (d) 1/2.

Explanation:

The question involves calculating probabilities from a special deck of 16 cards with four different colors and numbers.

a. Probability of Drawing a Two or a Four

Since there are 4 twos and 4 fours in the deck, the total number of favorable outcomes is 4 + 4 = 8. The total number of cards is 16. Thus, the probability of drawing a two or a four is 8/16, which simplifies to 1/2 or 0.5.

b. Probability Given that the Card is Not a One

If we know the card is not a one, we exclude the four ones from consideration, leaving us with 12 cards. Out of these, there are still 8 cards that are either a two or a four. Therefore, the probability is 8/12, which simplifies to 2/3.

c. Probability Given that the Card is a Two or a Three

Given that the card is a two or a three, there are 4 twos and 4 threes, so 8 possible cards. Out of these, 4 are twos or fours, so the probability is 4/8, which simplifies to 1/2 or 0.5.

d. Probability Given that the Card is Red or Green

Given that the card is red or green, there are 4 red and 4 green cards, so 8 possible cards. Out of these, 2 are twos and 2 are fours, leading to 4 favorable outcomes. The probability is therefore 4/8, which simplifies to 1/2 or 0.5.

Answer:

a) [tex] \bar X = 52[/tex]

[tex] Median = 55[/tex]

b) [tex]Q_1= \frac{44+44}{2}=44[/tex]

[tex]Q_3= \frac{57+60}{2}=58.5[/tex]

c) [tex] Range = Max -Min = 69-36=33[/tex]

d) [tex] s^2 =100.1429[/tex]

[tex] s= \sqrt{100.143}=10.0071[/tex]

e) [tex] Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25[/tex]

[tex] Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75[/tex]

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

Step-by-step explanation:

For this case we have the following dataset:

55 56 44 43 44 56 60 62 57 45 36 38 50 69 65

A total of 15 observations

Part a

We calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^{15} X_i}{15}[/tex]

And for this case we got [tex] \bar X = 52[/tex]

For the median we ust need to order the data on increasing way like this:

36, 38,43,44,44,45,50,55,56,56,57,60,62,65,69

Since the number of observations is an odd number the median would be on the 8 position from the dataset ordered on this case:

[tex] Median = 55[/tex]

Part b

In order to calculate the Q1 we need to select the following data:

36, 38,43,44,44,45,50,55

And the Q1 would be the average between the 4 and 5 positions like this:

[tex]Q_1= \frac{44+44}{2}=44[/tex]

And for the Q3 we select these values:

55,56,56,57,60,62,65,69

And the Q3 would be the average between the 4 and 5 positions like this:

[tex]Q_3= \frac{57+60}{2}=58.5[/tex]

Part c

The Range is defined as:

[tex] Range = Max -Min = 69-36=33[/tex]

Part d

In order to calculate the sample variance we can use the following formula:

[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And if we replace we got:

[tex] s^2 =100.1429[/tex]

And the deviation is just the square root of the variance:

[tex] s= \sqrt{100.143}=10.0071[/tex]

Part e

For this case we need to find the lower and upper limits for the boxplot given by:

[tex] Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25[/tex]

[tex] Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75[/tex]

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

Answer:

[tex]P(t) = 880(1.4142)^{t}[/tex]

Step-by-step explanation:

The bacteria's population may be expressed by the following equation:

[tex]P(t) = P_{0}(1+r)^{t}[/tex]

In which [tex]P_{0}[/tex] is the initial population, and r is the growth rate, as a decimal.

A bacteria culture starts with 880 bacteria and grows at a rate proportional to its size. After 4 hours there will be 3520 bacteria.

This means that [tex]P_{0} = 880, P(4) = 3520[/tex].

(a) Express the population P after t hours as a function of t . Be sure to keep at least 4 significant figures on the growth rate.

We have to find the growth rate, which we do applying the value of P(4) to the equation.

[tex]P(t) = P_{0}(1+r)^{t}[/tex]

[tex]3520 = 880(1+r)^{4}[/tex]

[tex](1+r)^{4} = \frac{3520}{880}[/tex]

[tex](1+r)^{4} = 4[/tex]

Applying the fourth root to both sides of the equality.

[tex]1 + r = 1.4142[/tex]

[tex]r = 0.4142[/tex]

So the equation of P(t) is:

[tex]P(t) = 880(1.4142)^{t}[/tex]

The population P after t hours for a bacteria culture growing at a rate proportional to its size is given by the exponential growth formula P(t) = 880 * e^((ln(4)/4)t), with an initial population of 880 and a growth rate determined from the population size after 4 hours.

The student is asking for an expression of the population P after t hours for bacteria that grow at a rate proportional to their size. Since the bacteria population growth is exponential, we use the formula P(t) = P0 * e^(rt), where P0 is the initial amount of bacteria, r is the growth rate, and t is time in hours.

We are given an initial population of 880 bacteria (P0) and after 4 hours we have 3520 bacteria. Using this information, we can find the growth rate r. The formula with the known values plugged in is 3520 = 880 * e^(4r). To solve for r, we first divide both sides by 880, getting 4 = e^(4r), and then take the natural logarithm of both sides to get ln(4) = 4r. Thus r = (ln(4))/4.

Now, the formula for P after t hours can be expressed as P(t) = 880 * e^((ln(4)/4)t). This formula will give the size of the bacteria population at any time t, in hours, assuming a constant, exponential growth rate.

Answer:

Explanation:

Question 1. Between what two values will approximately 68% of the numbers of days be?.

You can answer based on the 68-95-99.7% rule. As per this rule, about 68% of the data of a normal distribution (bell shaped) are within one standard deviation of the mean.

Here the mean is 32 day and the standard deviation is 7 day. Then 68% are in the interval 32 days ± 7 days.

That is:

Consequently, approximately 68% of the numbers of days will be between 25 and 39 days.

Question 2. Estimate the percentage of customer accounts for which the number of days is between 18 and 46.

First must determine the Z-scores both both values X = 18 and X = 46

The formula is:

         [tex]Z-score = (X-mean)/(standard\text{ }deviation)[/tex]

        [tex]Z-score=(18-32)/7=-2[/tex]

       [tex]Z-score=(46-32)/7=2[/tex]

Hence, you want to estimate the percentage of customers accounts for which the the number of days is within 2 standard deviations of the mean.

As per the 68-95-99.7 rule about 95% of the data are within 2 standard deviations of the mean. You can calculate it also from a standard normal distribution table, finding the area to the left of Z-score = - 2 and subtracting the area to the right of Z-score equal to 2: That is: 0.9772 - 0.0228 = 0.9484 = 95.44% ≈ 95%.

Question 3. Estimate the percentage of customer accounts for which the number of days is between 11 and 53.

Again, determine the Z-scores for the two values, X = 11 and X = 53.

         [tex]Z-score=(11-32)/7=-3[/tex]

        [tex]Z-score=(53-32)/7=3[/tex]

Hence, you want to estimate the probability of the number of days s between - 3 and 3 standard deviations.

Such probability is about 99.7%, according to the 68-95-99.7 rule.

If you use a standard distritution table you will find that the area to the right of the Z-score of -3 is 0.99865, thus the probability of the Z-score be to the right of 3 is 1 - 0.99865 = 0.00135.

And the probability in between -3 and 3 standard deviations is 0.99865 - 0.00135 = 0.9973 = 99.73% ≈ 99.7%.

Estimate the percentage of bills for number 39 when a bill was sent and when payment was made Answer:

Step-by-step explanation: yes