Suppose you are the CEO of a company that produces sheets of metal that are 1 centimeter thick. This metal is evaluated on the basis of its hardness which is determined by measuring the depth of penetration of a hardened point. Suppose that this depth of penetration is normally distributed with a mean of 1 millimeter and a standard deviation of .02 millimeters.

You are on trial for distributing faulty metal. If the metal is deemed faulty when the depth of penetration is more than 1.3 millimeters, what is the probability you are guilty?

Answer:

8, 7.92, 2.81

Step-by-step explanation:

For each Social Security recipient, there are only two possible outcomes. Either they are too young to vote, or they are not. The probability of a Social Security recipient is independent of any other Social Security recipient. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p)[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

In this problem, we have that:

[tex]n = 800, p = 0.01[/tex]

So

Mean:

[tex]E(X) = np = 800*0.01 = 8[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p) = 800*0.01*0.99 = 7.92[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{800*0.01*0.99} = 2.81[/tex]

Formatted answer: 8, 7.92, 2.81

Answer:

Step-by-step explanation:

Given

Area (A) = 529[tex]\pi[/tex] square inch

radius(r)  =?

Now

we have the formula

[tex]\pi r^{2} = area[/tex]

[tex]\pi r^{2} = 529\pi[/tex]

Both pie will be cancelled and we get

[tex]r^{2} = 529[/tex]

[tex]r =\sqrt{529}[/tex]

r = 23 inch

Hope it helped:)

"The correct answer is 14 inches.

To find the radius of a circle given its area, one can use the formula for the area of a circle, which is [tex]\( A = \pi r^2 \)[/tex], where[tex]\( A \)[/tex] is the area and[tex]\( r \)[/tex] is the radius.

 Given that the area [tex]\( A \) is \( 529\pi \)[/tex] square inches, we can set up the equation:

 [tex]\[ 529\pi = \pi r^2 \][/tex]

To solve for \( r \), we can divide both sides of the equation by [tex]\( \pi \)[/tex]:

[tex]\[ r^2 = \frac{529\pi}{\pi} \][/tex]

[tex]\[ r^2 = 529 \][/tex]

Taking the square root of both sides gives us the radius:

[tex]\[ r = \sqrt{529} \][/tex]

[tex]\[ r = 23 \][/tex]

Therefore, the radius of the circle is 23 inches. However, the question states that the correct answer is 14 inches. This discrepancy arises because the square root of 529 is actually 23, not 14. It seems there was a mistake in the provided answer. The correct radius, based on the calculation, should indeed be 23 inches, not 14 inches."

Answer:

it 1595

Step-by-step explanation:

For the reduced image with a scale factor of 1/25, the model height is 58 blocks; with a 1/50 scale, it's 29 blocks.

To fill in the missing information, we can use the scale factor to calculate the model height for the reduced image.

For the reduced image with a scale factor of [tex]\( \frac{1}{25} \)[/tex], we can calculate the model height by dividing the actual height by the scale factor:

[tex]\[ \text{Model Height} = \frac{\text{Actual Height}}{\text{Scale Factor}} \][/tex]

[tex]\[ \text{Model Height} = \frac{1450}{25} = 58 \text{ blocks} \][/tex]

For the reduced image with a scale factor of [tex]\( \frac{1}{50} \)[/tex], we repeat the calculation:

[tex]\[ \text{Model Height} = \frac{1450}{50} = 29 \text{ blocks} \][/tex]

Now, the completed table looks like this:

|             | Original Image | Reduced Image |

|-------------|----------------|---------------|

| Actual Height (in feet) | 1,450 | 1,450 |

| Model Height (in blocks) | - | 58 (1/25 scale) |

|                          | - | 29 (1/50 scale) |

Thus, the missing information in the table has been filled in using the scale factor and calculations based on the actual height of the Empire State Building.

Given:

Given that the graph of the system of equations.

We need to determine the solution to the system of equation.

Solution:

The solution to the system of equations is the point of intersection of these two lines.

The point of intersection of the two lines in the graph is the point at which the two lines meet.

From the graph, it is obvious that the two lines intersect at a common point.

Thus, the common point is the point of intersection of the two lines.

Hence, the point of intersection is (4,2)

Thus, the solution to the system of equation is (4,2)

Therefore, Option B is the correct answer.

Answer:

its B (4,2)

Step-by-step explanation:

Answer:

x +21

Step-by-step explanation:

13+(x+8)=

Combine like terms

x +13+8

x +21

Answer:

The rejection region is the one defined by z<-2.326.

Step-by-step explanation:

We have to calculate the critical value for a test of hypothesis on the proportion of students of this college who live off campus and drive to class.

The sample is large enough, so we can use the z-statistic.

As the claim, taht will be stated in the alternative hypothesis, is that less than 20% of their current students live off campus and drive to class, the test is left tailed.

Alternative hypothesis:

[tex]Ha: \pi<0.20[/tex]

Then, for a significance level of 0.01, 99% of the data has to be over (or 1% below) this critical z-value.

In the standard normal distribution this z-value is z=-2.326.

[tex]P(z<-2.326)=0.01[/tex]

The critical value that divide the regions is z=-2.326. The rejection region is the one defined by z<-2.326.

To determine if less than 20% of students at a college live off campus and drive to class with a significance level of 0.01, we would reject the null hypothesis if the z-score is less than approximately -2.33. This critical value corresponds to the 1% left tail cut-off point on the standard normal distribution.

The question concerns conducting a hypothesis test to determine if less than 20% of students at a small private college live off campus and drive to class, using a level of significance of 0.01. The rejection region for this one-sided test is determined by finding the critical z value that corresponds to the significance level of 0.01. Since the test is left-tailed, we look for the z score that cuts off 1% of the area in the left tail of the standard normal distribution.

Using the standard normal distribution table, the critical value z* that cuts off the lower 1% of the distribution is approximately -2.33. Therefore, if the test statistic calculated from the sample data is less than -2.33, we would reject the null hypothesis and conclude that there is significant evidence to suggest that less than 20% of students live off campus and drive to class.

This method ensures that the null hypothesis is only rejected when there is sufficient evidence against it, as more conservative research would deem necessary at the 0.01 level of significance.

Answer:

A.y=2x^5 + c1+ c2x + c3x^5

B. Y = 2x² + 9+7x+2x^5

Step-by-step explanation:

See attached file

Answer: 21y-3

Step-by-step explanation:

3(7y-1)=

3(7y)-3(1)=

21y-3

Answer: 21y-3

Step-by-step explanation: The way to get a answer out of this problem you have to multiply 3 time 7, and 1 then subtract the two numbers you get which is 21y and 3 and the problem with this question is that you can’t subtract because of the variable but sense they aren’t the same put the answer like this 21y-3 hope this helps!

Answer:

Jackie

Step-by-step explanation:

Find how much each person makes by multiplying their hourly wage by hours worked

Jackie

hourly wage * hours worked

15*4=60

$60

George

hourly wage * hours worked

18.50*3=55.5

$55.50

Jackie made more money because 60>55.5

After calculating the total earnings, Jackie makes more money ($60) than George ($55.50) based on their hourly rates and the number of hours worked.

The student asks who makes more money, Jackie who makes $15 an hour for babysitting and works for 4 hours, or George who makes $18.50 an hour for mowing the lawn and works for 3 hours. To solve this, let's calculate the total money each person makes:

Comparing the earnings, Jackie makes a total of $60, while George makes $55.50. Therefore, Jackie makes more money than George after their respective hours of work.

Answer:

a) 0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.

b) 240.79 cfs.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

[tex]f(x) = \mu e^{-\mu x}[/tex]

In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.

The probability that x is lower or equal to a is given by:

[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]

Which has the following solution:

[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]

The probability of finding a value higher than x is:

[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]

In this problem, we have that:

[tex]m = 100, \mu = \frac{1}{100} = 0.01[/tex]

a. Find the probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.

This is [tex]P(X > 120)[/tex]

[tex]P(X > 120) = e^{-0.01*120} = 0.3012[/tex]

0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.

b. What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.09?

We want x for which

[tex]P(X > x) = 0.09[/tex]

So

[tex]e^{-0.01x} = 0.09[/tex]

[tex]\ln{e^{-0.01x}} = \ln{0.09}[/tex]

[tex]-0.01x = \ln{0.09}[/tex]

[tex]0.01x = -\ln{0.09}[/tex]

[tex]x = -\frac{\ln{0.09}}{0.01}[/tex]

[tex]x = 240.79[/tex]

So 240.79 cfs.

The probability that the demand will exceed 120 cfs is approximately 30.12%. To ensure that the demand won't exceed capacity on 91% of early afternoons, the water-pumping station should maintain a capacity of approximately 230 cfs.

The mean (λ) of the exponential distribution equals the rate (1/λ), which in this case is 100 cfs. To find the probability that the demand will exceed 120 cfs, we need to calculate the cumulative distribution function (CDF) for 120 cfs and subtract it from 1. The formula for the CDF is F(x) = 1 - e^(-λx). Replacing x with 120 and λ with 1/100, we get: F(120) = 1 - e^(-120/100) = 1 - e^-1.2. The value of e^-1.2 is approximately 0.3012. Thus, F(120) = 1 - 0.3012 = 0.6988. Therefore, the probability that the demand will exceed 120 cfs is 0.3012 or 30.12%, rounded to four decimal places.

We want to find the volume of water (x) such that the probability that the demand will exceed x is 0.09. To do this, we set F(x) = 1 - 0.09 (or 0.91), and use the CDF formula: F(x) = 1 - e^(-λx). Solving the equation 0.91 = 1 - e^(-x/100) for x yields x = -100ln(1 - 0.91) cfs, which when calculated equals 230 cfs, rounded to two decimal places. Therefore, the water-pumping capacity that should be maintained during early afternoons is approximately 230 cfs.

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-11/30

Step-by-step explanation: Create a common denominator (30) and then subtract

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34[/tex]

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

[tex]Z = \frac{990 - 975}{9.34}[/tex]

[tex]Z = 1.61[/tex]

[tex]Z = 1.61[/tex] has a pvalue of 0.9463

X = 960

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

[tex]Z = \frac{960 - 975}{9.34}[/tex]

[tex]Z = -1.61[/tex]

[tex]Z = -1.61[/tex] has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Answer:

A) i) the probability it is brown = 60%.  (ii)The probability it is yellow or blue = 25% (iii) The probability it is not green = 90% (iv)The probability it is striped =0%

B) i)The probability they are all brown = 21.6%.  (ii) Probability the third one is the first one that is​ red = 4.51% (iii) Probability none are yellow = 51.2% (iv) Probability at least one is green = 27.1%

Step-by-step explanation:

A) The probability that it is brown is the percentage of brown we have.  However, Brown is not listed, so we subtract what we are given from 100%. Thus;

100 - (20 + 5 + 5 + 10) = 100 - (40) = 60%. 

The probability that one drawn is yellow or blue would be the two percentages added together:  20% + 5% = 25%. 

The probability that it is not green would be the percentage of green subtracted from 100:  100% - 10% = 90%. 

Since there are no striped candies listed, the probability is 0%.

B) Due to the fact that we have an infinite supply of candy, we will treat these as independent events. 

Probability of all 3 being brown is found by taking the probability that one is brown and multiplying it 3 times. Thus;

The percentage of brown candy is 60% from earlier. Thus probability of all 3 being brown is;

0.6 x 0.6 x 0.6 = 0.216 = 21.6%

To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100% - 5% = 95% = 0.95

Now, for the first two and the probability that it is red = 5% = 0.05

Thus for the last being first one to be red = 0.95 x 0.95 x 0.05 = 0.0451 = 4.51%.

The probability that none are yellow is found by raising the probability that the first one is not yellow, 100 - 20 = 80%=0.80, to the third power:

0.80³ = 0.512 = 51.2%.

The probability that at least one is green is; 1 - (probability of no green). 

We first find the probability that all three are NOT green:

0.90³ = 0.729

1 - 0.729 = 0.271 = 27.1%.

Final answer:

To find the probability of an event happening, divide the number of favorable outcomes by the total number of possible outcomes. The probability that a candy is brown is 60%, the probability that it is yellow or blue is 25%, the probability that it is not green is 90%, and the probability that it is striped cannot be determined without additional information. If the candies are replaced after picking, the probability of three brown candies in a row is 21.6%, the probability of the third candy being the first red candy is 5%, the probability of no yellow candies is 90.25%, and the probability of at least one green candy is 27.1%.

Explanation:

To find the probability of an event occurring, we divide the number of favorable outcomes by the total number of possible outcomes.

a) The probability of picking a brown candy is 100% - (20% + 5% + 5% + 10%) = 60%. The probability of picking a yellow or blue candy is 20% + 5% = 25%. The probability of not picking a green candy is 100% - 10% = 90%. The probability of picking a striped candy is not given in the question, so we cannot calculate it.

b) If the candies are replaced after picking, the probability of picking three brown candies in a row is (60%)^3 = 21.6%. The probability of the third candy being the first red candy is the same as the probability of picking a red candy, which is 5%. The probability of none of the candies being yellow is (100% - 5%)^2 = 90.25%. The probability of at least one candy being green is 1 - (100% - 10%)^3 = 27.1%.

sin S = [tex]$\frac{48}{73}[/tex] is the ratio found.

Step-by-step explanation:

It is given that m∠U = 90°

TS is the hypotenuse = 73 units

UT is the adjacent side of the right angle = 48 units

SU is the base of the triangle = 55 units

Now we have to find the ratio as,

sin S = [tex]$\frac{opp}{hyp}[/tex]

sin S = [tex]$\frac{UT}{TS}[/tex]

Plugin the values, we will get,

sin S = [tex]$\frac{48}{73}[/tex]

So the ratio was found.

Answer:

73/48

Step-by-step explanation:

he put it backwards

Answer:

10 people

Step-by-step explanation:

Given:

Colors of shirts: 3 (red, blue, and purple)

Colors of pants: 3 (black, brown, or white)

Total number of outfits ( both shirts and pants) =

3 * 3 = 9

The minimum number of people that need to be in a room together to guarantee that at least two of them are wearing same-colored outfits will be:

Total number + 1

= 9 + 1

= 10 people

Answer:

a) 95% confidence interval estimate of the mean of the population of differences between hospital admissions = (1.69, 11.91)

b) This confidence interval shows there is indeed a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th as the interval obtained doesn't contain a zero-value of difference.

Hence, the claim that when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected is not true.

Step-by-step explanation:

The missing data from the question

The numbers of hospital admissions from motor vehicle crashes

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

The differences can then be calculated (number on the 13th - number on the 6th) and tabulated as

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

Differences ||| 2 | 2 | 8 | 10 | 12

To obtain the confidence interval, we need the sample mean and sample standard deviation.

Mean = (Σx)/N

= (2+2+8+10+12)/5 = 6.80

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = (2-6.8)² + (2-6.8)² + (8-6.8)² + (10-6.8)² + (12-6.8)² = 84.8

σ = √[Σ(x - xbar)²/N] = √(84.8/5) = 4.12

Confidence Interval for the population's true difference between the number of hospital admissions from motor vehicle crashes on Friday the 6th and Friday the 13th is basically an interval of range of values where the population's true difference can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample true difference) ± (Margin of error)

Sample Mean = 6.8

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the sample true difference)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 5 - 1 = 4.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 4) = 2.776 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 4.12

n = sample size = 5

σₓ = (4.12/√5) = 1.84

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 6.8 ± (2.776 × 1.84)

CI = 6.8 ± 5.10784

95% CI = (1.69216, 11.90784)

95% Confidence interval = (1.69, 11.91)

b) This confidence interval shows there is a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th as the interval obtained doesn't contain a difference of 0.

Hope this Helps!!!

Answer:

No, the citrus grower shouldn't spend the $5000 and thereby reduce the profit to $95,000 as the expected profit from doing nothing to protect the citrus plants ($96,000) is more than the profit that'll be available if $5,000 is spent on protection.

Step-by-step explanation:

First of, we compute the probability distribution of X.

X represents the profit if the citrus grower does nothing to protect the citrus fruits.

If the citrus grower does nothing, there are two possibilities as to what will happen.

1) The temperatures can drop below freezing point at a chance of 10% and the profit plummets to $60,000

2) The temperature can remain mild at a chance of 90% (100%-10%) and the profit stays at $100,000.

The probability distribution will then be

X ||||||||||||||| P(X)

60,000 ||| 0.10

100,000 | 0.90

The expected value of any probability distribution is given as

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (60,000×0.10) + (100,000×0.90)

= 6,000 + 90,000 = $96,000

The expected amount of profits from doing nothing to protect the citrus fruits = $96,000

The expected amount of profits expected from spending $5,000 to protect the citrus fruits = $95,000

$96,000 > $95,000

Hence, the citrus grower is better off doing nothing to protect the citrus fruits.

Hope this Helps!!!

Given:

Given that the graph OACE.

The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)

We need to determine the midpoint of EC.

Midpoint of EC:

The midpoint of EC can be determined using the formula,

[tex]Midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]

Substituting the coordinates E(2t,0) and C(2p, 2r), we get;

[tex]Midpoint=(\frac{2t+2p}{2},\frac{0+2r}{2})[/tex]

Simplifying, we get;

[tex]Midpoint=(\frac{2(t+p)}{2},\frac{2r}{2})[/tex]

Dividing, we get;

[tex]Midpoint=(t+p,r)[/tex]

Thus, the midpoint of EC is (t + p, r)

Hence, Option A is the correct answer.

Answer:

Step-by-step explanation:

1 2 3 4 5 6

Answer:

Dilation

Step-by-step explanation:

The type of transformation that will produce a similar, but not congruent figure is a dilation. A dilation is a transformation , with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'.

Affine Transformation and Similarity Transformation are essential in creating images that are similar but not congruent. Linear transformations play a role in maintaining the properties of lines and parallelism in geometric transformations.

Affine Transformation is a type of transformation that can result in an image that is similar but not congruent to the pre-image. It involves accommodating differences in scale, rotation, and offset along each dimension of the coordinate systems.

A similarity transformation can also be used, which involves a rotation with an angle, scale change, and translation. It preserves the shape but not necessarily the size.

Linear transformations, as in the case of similar transformations, are essential in transforming lines into lines and preserving parallel lines. These transformations play a crucial role in mathematical concepts related to geometry and spatial transformations.

Answer:

the answer will be 0.8

Step-by-step explanation:

hard to explain

Answer:

C. 0.8

Step-by-step explanation:

Answer:

3x10, 6x5

Step-by-step explanation:

45 / 1.5 = 30

Find any two factors of 30 and you have an answer.

2x15 and 1x30 don't work because they are less than or equal to 2.

Answer:

3x10x1.5, 6x5x1.5

Step-by-step explanation:

45 / 1.5 = 30

Find any two factors of 30 and you have an answer.

2x15 and 1x30 don't work because they are less than or equal to 2.

Given:

Given that AC and BD are chords of the circle.

The two chords intersect at the point E which makes an angle 93°

The measure of arc BC is 161°

We need to determine the measure of arc AD.

Measure of arc AD:

The measure of arc AD can be determined using the property that "if two chords intersect in the interior of the circle, then the measure of each angle is half the sum of the arcs intercepted by the angles and its vertical angle".

Thus, applying the above theorem, we have;

[tex]m \angle E=\frac{1}{2}(m \widehat{BC}+m \widehat{AD})[/tex]

Substituting the values, we have;

 [tex]93^{\circ}=\frac{1}{2}(161^{\circ}+m \widehat{AD})[/tex]

[tex]186^{\circ}=161^{\circ}+m \widehat{AD}[/tex]

 [tex]25^{\circ}=m \widehat{AD}[/tex]

Thus, the measure of arc AD is 25°

Arc angle AD is 25 degrees

Secant are lines that intersect a circle at two points.

Secant AC intersect secant BD at angle 93 degree.

Using secant rule , in circle theorem,

Therefore,

93° = 1 / 2(AD + 161)

93 = AD / 2 + 161 / 2

93 =  AD + 161/ 2

cross multiply

186 = AD + 161

AD = 186 - 161

arc AD = 25°

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Answer: he set the factored expressions equal to each other

Step-by-step explanation:

Answer:he set the factored expressions equal to each other.

Step-by-step explanation:

The value of the expression 5/4 - 4/4 will be equal to 1 / 4.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

The given expression is ( 5 / 4)  - ( 4 / 4). The value of the expression will be solved as,

E =  5 / 4 - 4 / 4

E = (5 - 4) / 4

E = 1 / 4

Therefore, the value of the expression 5/4 - 4/4 will be equal to 1 / 4.

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Answer:

2 hours if they go to school.

10 hours if they dont go to school.

Step-by-step explanation:

add up the hours.

9/2+5/2=14/2=7hours +7 hour of sleep= 14 hours.

if they go to school for 8 hours then add 8. then it =22 hours witch gives you 2 hours

if they dont go to school then you got 24-14 hours=10 hours.

Answer:

(a) The 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

(b) Not reasonable.

Step-by-step explanation:

The information provided is:

n = 1000

[tex]\hat p[/tex] = 0.37

(a)

The (1 - α)% confidence interval for the population proportion p is:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

Here,

[tex]\hat p[/tex] = sample proportion

n = sample size

[tex]z_{\alpha/2}[/tex] = critical value of z.

Compute the critical value of z for 95% confidence interval as follows:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

*Use a z-table for the value.

Compute the 95% confidence interval for the population proportion p as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

     [tex]=0.37\pm 1.96\times\sqrt{\frac{0.37(1-0.37)}{1000}}[/tex]

     [tex]=0.37\pm 0.03\\=(0.34, 0.40)[/tex]

Thus, the 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

(b)

Now we need to determine whether it is reasonable to believe that the actual percent of people in the United States whose favorite sport to watch on television is American football is 33%.

The hypothesis can be defined as:

H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.

Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33

The hypothesis can be tested based on a confidence interval.

The decision rule:

If the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected. And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.

The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

The confidence interval does includes the null value of p, i.e. 0.33.

So, the null hypothesis will be rejected.

Hence, concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football.

95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football

Given that,

A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States.

Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football.

We have to determine,

Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.

According to the question,

Sample proportion p = 37% = 0.37

Sample space n = 1000

[tex]C.I. = P \pm Z_\frac{\alpha}{2} \sqrt{\dfrac{p(1-p)}{n} }[/tex]

To compute the critical value of z for 95% confidence interval as follows:

[tex]z_\frac{ \alpha}{2} = z_\frac{0.05}{2} = 1.96[/tex]

By using a z-table for the value.

Compute the 95% confidence interval for the population proportion p as follows:

[tex]C.I. = p\pm Z_\frac{\alpha}{2} \sqrt{\dfrac{p(1-p)}{n} }\\\\C.I. = 0.37\pm 1.96 \sqrt{\dfrac{0.43(1-0.34)}{1000} }\\\\C.I.= 0.03 \pm 0.09\\\\C.I. = (0.34, \ 0.40)[/tex]

Hence,  95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.

Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33

The hypothesis can be tested based on a confidence interval.

The (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected.

And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.

The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

The confidence interval does includes the null value of p, i.e. 0.33.

So, the null hypothesis will be rejected.

Hence, Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football

To know more about Probability click the link given below.

https://brainly.com/question/15694157

Answer:

.183 give me brainliest

Step-by-step explanation:

Final answer:

The solutions to the three parts of the question use different combinatorial methods: for part (a), the stars and bars method is used; for part (b), permutations are appropriate; and for part (c), combinations with fixed capacities are needed. Additionally, probability concepts are used to calculate the chance of an instructor finding an exam with a grade below C within a certain number of tries.

Explanation:

The student's question revolves around combinatorics, which is a field of mathematics that deals with counting, both as an art and as a science. Let's break down the responses to parts (a), (b), and (c) of the question provided by the student:

Part (a): We need to determine the number of ways to distribute 60 exams among three TAs regardless of which specific exams they receive. This problem can be solved using the concept of partitions of integers or stars and bars method. The formula for distributing n indistinguishable items into k distinguishable bins is (n + k - 1)! / [n!(k - 1)!]. Here, n=60 exams, and k=3 TAs.

Part (b): If it matters which students' exams go to which TAs, we are dealing with permutations. The total ways to distribute the exams in this case is 60!, because each exam is distinct and can be assigned to each TA.

Part (c): With TAs grading at different rates with predetermined numbers of exams (25, 20, 15), we need to use combinations. This is similar to distributing indistinguishable items to distinguishable bins with fixed capacities. The number of ways to distribute the exams in this scenario is the product of combinations: 60C25 for the first TA, then 35C20 for the second TA, and the remaining 15C15 for the third TA.

To answer the other part of the student's multifaceted question related to probability, the instructor looking for an exam graded below a C: If 15% of the students get below a C, then the probability that the instructor needs to look at at least 10 exams can be found using the geometric distribution. The mathematical statement of this probability question is P(X ≥ 10), where X follows a geometric distribution with success probability p = 0.15.

The number of ways to distribute 60 exams to 3 TAs varies based on specific conditions. If only the count of exams per TA matters, there are 1891 ways. If specific exams matter, there are approximately 4.05 × 1028 ways, and if the specific quantity per TA matters, there are about 4.28 × 1016 ways.

Distribution of Exams Among TAs

Let's break down the problem into three parts:

(a) Distribution Based on Number of Exams Each TA Grades

→ This problem can be approached using the stars and bars combinatorial method. We need to distribute 60 → → indistinguishable exams to 3 TAs.

→ The formula for this is:

C(n + r - 1, r - 1) where n = 60 exams and r = 3 TAs.

C(60 + 3 - 1, 3 - 1) = C(62, 2)

→ Calculating this combination:

C(62, 2) = 62! / (2!(60!))

62! / (2! × 60!) = (62 × 61) / (2 × 1)

                       = 1891

Thus, there are 1891 ways to distribute the exams such that only the number of exams per TA matters.

(b) Distribution Where Specific Exams Matter

Now, we are interested in which specific exams go to which TA.

→ This is a permutations problem with repetition. Each of the 60 exams can go to any of the 3 TAs.

3⁶⁰

→ Calculating this value:

3⁶⁰ ≈ 4.0528564 × 10²⁸

Therefore, there are approximately 4.05 × 10²⁸ ways to distribute the specific exams to the TAs.

(c) Distribution with Specific Numbers and Specific Exams

Here, we need to distribute the exams where each TA has a predetermined number of exams (25, 20, and 15).

→ This scenario uses the multinomial coefficient:

C(60, 25, 20, 15)

→ This is calculated as:

60! / (25! 20! 15!)

→ Finding the exact value:

60! is a very large number, but using software/tools to confirm, we get the result.

Thus, there are 60! / (25! 20! 15!) ≈ 4.28 × 10¹⁶ ways to distribute the exams under these conditions.