It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

Answer:

175$

Step-by-step explanation:

Given that Pam  is an employee at a jewelry kiosk in a mall. If Pam works hard, there is a 75% probability that jewelry profits will equal $400 a day and a 25% probability that jewelry profits will equal $200 a day. If Pam shirks, there is a 25% probability that jewelry profits will equal $400 a day and a 75% probability that jewelry profits will equal $200 a day.

Since Pam works hard

expected value of profit for the kiosk = [tex]400(0.75)+200(0.25)\\=350[/tex]

From this value, 50% would be given to Pam

Hence expected gain for Pam for working hard

= [tex]0.5(350)\\=175[/tex]

175 $

The population of the country decreased by approximately 2.4% between 1990 and 2010, calculated using the percent decrease formula $(167 million - 163 million) / 167 million * 100$.

To find the percent decrease in the country's population, we first need to determine the actual decrease in population. In this case, the population in 1990 was about 167 million and it decreased to 163 million in 2010, so the numeric decrease is 167 million - 163 million = 4 million.

Next, we compute the percent decrease based on the initial number from 1990. The formula for percent decrease is: (Decrease in Value / Original Value) * 100. Therefore, the percent decrease is (4 million / 167 million) * 100 = 2.4% (rounded to one decimal place).

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The decrease in the country's population from 1990 to 2010 is approximately 2.4%.

To find the percentage decrease in the country's population, we first calculate the amount of decrease. This is done by subtracting the population in 2010 from the population in 1990, in other words, 167 million - 163 million, which equals 4 million. Next, we divide this number by the original population(the population in 1990) and times the result by 100%. Therefore, the percentage decrease in population is (4/167) * 100%, approximately equals 2.4% when rounded to one decimal place.

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Answer:

The original number is 10 times the new number.

Step-by-step explanation:

Answer:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

(–2975, 175)

Do not reject H₀

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the salaries of elementary school teachers are equal in two districts. You can test this trough the population means of the salaries of the teachers if they are either equal or different or directly test if the difference between the salaries of the two districts is cero or not, symbolically: μ₁ - μ₂ = 0

Remember, in the null hypothesis is usually stated the known information, is the "no change" premise and always carries the = symbol.

The hypothesis is:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

You have two normally distributed variables and you are studying the difference of the means. You can use a pooled Z to make the interval, the formula is:

X[bar]₁-X[bar]₂ ± [tex]Z_{1-\alpha /2}[/tex]*(√(σ₁²/n₁+σ₂²/n₂)

Since the test is two-tailed and at a signification level of 5% I've made the interval at the complementary confidence level of 95% so that I can use it to decide over the hypothesis.

X[bar]₁-X[bar]₂ ± [tex]Z_{1-\alpha /2}[/tex]*(√(σ₁²/n₁+σ₂²/n₂)

[tex]Z_{1-\alpha /2}[/tex] = [tex]Z_{0,975}[/tex] = 1,96

[28900-30300 ± 1.96*(√((2300)²/15+(2100)²/15)]

[-1400 ± 1576,15]

[-2976.15;176,15]

Since I've approximated to two decimal units in the intermediate calculations, the values ​​differ slightly, but the interval is:

(–2975, 175)Now, since the 0 is contained in the Confidence interval, the decision is to not reject the null hypothesis. In other words, the difference in average salaries between the two districts is cero.

I hope it helps!

The appropriate set of hypotheses (H0, H1) is B. µ1 – µ2 ≠ 0, µ1 – µ2 = 0.

The null hypothesis in a test predicts that there's no relationship between the variables. From the information, the null and alternative hypothesis are reflected by µ1 – µ2 ≠ 0, µ1 – µ2 = 0.

The confidence interval will be:

= (28900 - 30300) + 2.13 × ✓(2 × 4850000/15)

= = (28900 - 30300) + 2.13 × 804.15

= 325

Also, (28900 - 30300) - 2.13 × ✓(2 × 4850000/15)

= = (28900 - 30300) - 2.13 × 804.15

= -3125

The confidence interval will be (–3125, 325).

Lastly, the salaries of the teachers from the two districts isn't different as it's equal since the confidence interval has negative and positive values.

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Answer:

a) The distribution for the random variable X is given by:

X       |           -5         |    95        |      745      |        1995     |

P(X)   |  6992/7000  |  5/7000  |     2/7000 |       1/7000  |

b) E(X)=-4.43. That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43.

c) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]

Step-by-step explanation:

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.

Part a

The info given is:

N=7000 represent the number of tickets sold

$5 is the price for any ticket

Number of tickets with a prize of $2000 =1

Number of tickets with a prize of $750=2

Number of tickets with a prize of $100=5

Let X represent the random variable net gain when we buy an individual ticket. The possible values that X can assume are:

___________________________

Ticket price    Prize     Net gain (X)

___________________________

5                     2000       1995

5                     750          745

5                     100           95

5                      0              -5

___________________________

Now we can find the probability for each value of X

P(X=1995)=1/7000, since we ave just one prize of $2000

P(X=745)=2/7000, since we have two prizes of $750

P(X=95)=5/7000, since we have 5 prizes of $100

P(X=-5)=6992/7000. since we have 6992 prizes of $0.

So then the random variable is given by this table

X       |           -5         |    95        |      745      |        1995     |

P(X)   |  6992/7000  |  5/7000  |     2/7000 |       1/7000  |

Part b

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=(-5)*(\frac{6992}{7000})+(95)(\frac{5}{7000})+(745)(\frac{2}{7000})+(1995)(\frac{1}{7000})=\frac{-31000}{7000}=-4.43[/tex]

That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43.

Part c

In order to find the standard deviation we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=(25)*(\frac{6992}{7000})+(9025)(\frac{5}{7000})+(555025)(\frac{2}{7000})+(3980025)(\frac{1}{7000})=\frac{5310000}{7000}=758.571[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =758.571-(-4.43)^2 =738.947[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]

Answer:

t=5.86

We can conclude that the population correlation between their assets and pretax profit is higher than 0 at the significance level provided.

Step-by-step explanation:

n= 20 random sample taken

r=0.81 correlation coeffcient obtained

[tex]\alpha=0.025[/tex] significance level obtained

1) System of hypothesis

The system of hypothesis given are:

Null hypothesis :[tex]\rho \leq 0[/tex]

Alternative hypothesis: [tex] \rho >0[/tex]

2) Calculate the statistic

The statistic in order to test an hypothesis for the correlation coefficient is given by:

[tex]t =\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}[/tex]

This statistic follows a t distribution with n-2 degrees of freedom

If we replace the values given we got:

[tex]t =\frac{0.81\sqrt{20-2}}{\sqrt{1-(0.81^2)}}=5.86[/tex]

3) P value

For this case w eneed to calculate first the degrees of freedom

[tex]df=n-2=20-2=18[/tex]

And then analyzing the alternative hypothesis we can calculate the p value on this way:

[tex]p_v =P(t_{18} >5.86) =1-P(t_{18} <5.86)=1-0.99999=7.51x10^{-6}[/tex]

Since the P-value is smaller than the significance level, we have enough evidence to reject the null hypothesis in favor of the alternative. We conclude "there is sufficient evidence at the significance level to conclude that there is a linear relationship in the population between the two variables analyzed."

The decision rule for a 0.025 significance level intends to reject the null hypothesis if the p-value from the test statistic is less than 0.025. Using a standard Z test to compute the test statistic, we substitute the given values to find the result. As the p-value is 0.026, which is greater than the 0.025 significance level, we do not reject the null hypothesis, indicating insufficient evidence to conclude that the population correlation is greater than zero.

To answer your question, let's first state the decision rule for a 0.025 significance level. Considering your null hypothesis (H0: rho ≤ 0) and the alternative hypothesis (H1: rho > 0), we want to reject the null hypothesis if the p-value from the test statistic is less than 0.025.

Since we aren't given a specific formula or context about how the test statistic should be computed, I'll assume that it's via a standard Z test using the formula:  

where r is the sample correlation (0.81 in this case), and n is the number of data points (20 in this case). Let's substitute these values into the formula to find the test statistic.

Based upon the p-value from your computing software as 0.026 and the significance level being 0.025, we do not reject the null hypothesis because the p-value is greater than the significance level. In other words, there is insufficient evidence at the 0.025 significance level to conclude that the population correlation is greater than zero.

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Answer:

The sample should be 1,068.

Step-by-step explanation:

Consider the provided information.

Confidence level is 95% and margin of error is 0.03.

Thus,

1-α=0.95

α=0.05, E=0.03 and planning value [tex]\hat p=0.5[/tex]

Formula to calculate sample size is: [tex]n=\frac{\hat p(1-\hat p)(z_{\alpha/2})^2}{E^2}[/tex]

From the table we can find:  

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

Substitute the respective values in the above formula we get:

[tex]n=\frac{0.5(0.5)(1.96)^2}{(0.03)^2}[/tex]

[tex]n=\frac{0.25(1.96)^2}{(0.03)^2}\approx 1067.111[/tex]

Hence, the sample should be 1,068.

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

[tex]X_{IS}=63[/tex] number of high speed internet users that had been interrupted one or more times in the past month.

[tex]n=150[/tex] random sample taken

[tex]\hat p_{IS}=\frac{63}{150}=0.42[/tex] estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

[tex]p_{IS}[/tex] true population proportion of high speed internet users that had been interrupted one or more times in the past month.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

[tex]0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341[/tex]

[tex]0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499[/tex]

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58[/tex]

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

[tex]0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316[/tex]

[tex]0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524[/tex]

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32[/tex]

And rounded up we have that n=335

4) Part d

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599[/tex]

And rounded up we have that n=649

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean lead concentration of sea water samples

Number of samples. n = 6

Mean, u = 0.903 cc/cubic meter

Standard deviation, s = 0.0566

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

0.903 +/- 1.96 × 0.0566/√6

= 0.903 +/- 1.96 × 0.0566/2.44948974278

= 0.903 +/- 0.045

The lower end of the confidence interval is 0.903 - 0.045 =0.858

The upper end of the confidence interval is 0.903 + 0.045 =0.948

Therefore, with 95% confidence interval, the mean lead concentration of the sea water is between 0.858 cc/cubic meter and 0.948 cc/cubic meter

Answer:

Step-by-step explanation:

Perimeter of a rectangle is expressed as (2 length + 2 width) = 2(L + W)

A) The length of rectangle A is y + 1

The width of rectangle A is x

Perimeter of rectangle A = 2(y + 1 + x) = 2y + 2 + 2x

= 2x + 2y + 2

The length of rectangle B is 2x - 2y

The width of rectangle B is x + 1

Perimeter of rectangle B = 2(2x - 2y+ x + 1) = 4x - 4y + 2x + 2) = 4x + 2x - 4y + 2

= 6x - 4y + 2

The length of rectangle C is 3x + 3y

The width of rectangle C is 2x - 3

Perimeter of rectangle C = 2(3x + 3y + 2x - 3) = 6x + 6y + 4x - 6) =

(6x + 6y + 4x - 6)

= 10x + 6y - 6

B) The combined perimeters will be the sum if perimeter of rectangle A, perimeter of rectangle B and perimeter of rectangle C. It becomes

2x + 2y + 2 + 6x - 4y + 2 + 10x + 6y - 6

Collecting like terms

2x + 6x + 10x + 2y + 6y - 4y + 2 + 2 - 6

The combined perimeter = 18x + 4y - 2

Answer:

In bold below.

Step-by-step explanation:

A.

2x + 2(y + 1)

= 2x + 2y + 2.

2(2x - 2y) + 2(x + 1)

= 4x - 4y + 2x + 2

= 6x - 4y + 2.

2(3x + 3y ) + 2(2x - 3)

= 6x + 6y + 4x - 6

= 10x + 6y - 6.

B.

2x + 2y + 2 + 10x + 6y - 6 + 6x - 4y + 2

=  18x + 4y - 2.

Answer:

the probability is 0.28

Step-by-step explanation:

using Bayes's theorem

P(A|B)=P(A∩B)/P(B)

where

P(A∩B) = probability that events A and B happen

P(A|B) = probability that event A happen if B already happened

P(B)= probability of event B

therefore

P(A∩B)=P(A|B)*P(B)

if event A= selection of a student that lives in a dormitory  and event B = selection of a freshmen student

P(A|B) = 0.8 (live in a dormitory knowing that is a freshmen student )

P(B) = 0.35 (freshmen student)

P(A∩B)=P(A|B)*P(B) = 0.8* 0.35 =0.28

The probability that a randomly selected student is a freshman who lives in a dormitory is 0.28. So the correct option is  A.

To find the probability that a randomly selected student is a freshman who lives in a dormitory, we need to multiply the probability that a student is a freshman by the probability that a freshman lives in the dormitory. According to the given information, 35% of the students are freshmen, and among them, 80% live in the dormitory.

We calculate this probability using the formula:

Probability of (Freshman who lives in dorm) = Probability of (Freshman) × Probability of (Dorm | Freshman)

Probability of (Freshman who lives in dorm) = 0.35 × 0.80

Probability of (Freshman who lives in a dorm) = 0.28

Thus, the correct answer is A) 0.28.

Answer:

[tex]\lambda \geq 6.63835[/tex]

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that [tex]X \sim Poisson(\lambda)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

And f(x)=0 for other case.

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

[tex]P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}[/tex]

[tex]P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}[/tex]

And replacing we have this:

[tex]P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[][/tex]

[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)[/tex]

And we want this probability that at least of 99%, so we can set upt the following inequality:

[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99[/tex]

And now we can solve for [tex]\lambda[/tex]

[tex]0.01 \geq e^{-\lambda}(1+\lambda)[/tex]

Applying natural log on both sides we have:

[tex]ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)[/tex]

[tex]ln(0.01) \geq -\lambda+ln(1+\lambda)[/tex]

[tex]\lambda-ln(1+\lambda)+ln(0.01) \geq 0[/tex]

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

[tex]x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}[/tex]

Where :

[tex]f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)[/tex]

[tex]f'(x_n)=1-\frac{1}{1+\lambda}[/tex]

Iterating as shown on the figure attached we find a final solution given by:

[tex]\lambda \geq 6.63835[/tex]

The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.

This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.

Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:

P(X=k) = λ^k * exp(-λ) / k!

The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:

P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99

Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.

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Answer:

[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]

Δn=3

Step-by-step explanation:

Remember, if we need to divide the interval (a,b) in n equal subinterval, then we need divide the distance (d) between the endpoints of the interval and divide it by n. Then the width Δn of each subinterval is d/n.

We have the interval [-5,7]. The distance between the endpoints of the interval is

[tex]d=7-(-5)=12[/tex].

Now, we divide d by 4 and obtain [tex]\frac{d}{4}=\frac{12}{4}=3[/tex]

Then, Δn=3.

Now, to find the endpoints of each sub-interval, we add 3 from the left end of the interval.

[tex]-5=x_0\\x_0+3=-5+3=-2=x_1\\x_1+3=1=x_2\\x_2+3=4=x_3\\x_3+3=7=x_4[/tex]

So,

[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]

The width of each subinterval is 3. The endpoints in ascending order are -5, -2, 1, 4, 7.

To find the width of each subinterval, we need to divide the length of the interval by the number of subintervals. In this case, the interval is [-5,7] and we need to divide it into 4 equal subintervals. The length of the interval is 7 - (-5) = 12. So the width of each subinterval is 12/4 = 3.

Next, we can find the endpoints of the subintervals. Since we have 4 subintervals, we need 5 endpoints. The first endpoint is the left endpoint of the interval, which is -5. Then we add the width of each subinterval to find the next endpoints: -5 + 3 = -2, -2 + 3 = 1, 1 + 3 = 4, and finally 4 + 3 = 7, which is the right endpoint of the interval. Therefore, the endpoints in ascending order are -5, -2, 1, 4, 7.

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Answer:

16

Step-by-step explanation:

Let,

P = movie rated​ PG-13,

P' = movie does not rated PG-13,

E = movie earned over​ $100 million,

E' = movie earned less than $100 million,

According to the question,

[tex]n(P\cap E)=16[/tex]

[tex]n(P\cap E')=29[/tex]

[tex]n(P'\cap E')=39[/tex]

Since,

[tex]n(P\cap E)+n(P\cap E')+n(P'\cap E')+n(P'\cap E)[/tex] = total movies

[tex]16 + 29 + 39 + n(P'\cap E) = 100[/tex]

[tex]n(P'\cap E) = 100 - 84 = 16[/tex]

Hence,

The number of movies that earned over​ $100 million and that were not rated​ PG-13 would be 16.

To find the number of movies that earned over $100 million but were not rated PG-13, subtract the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million.

To find the number of movies that earned over $100 million but were not rated PG-13, we need to subtract the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million.

Substituting the given values, we have:

Total movies earning over $100 million - 16 = Movies not rated PG-13 earning over $100 million.

Therefore, the number of movies that earned over $100 million and were not rated PG-13 is 100 - 16 = 84.

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To determine the number of cards each class parent helper will receive, divide the total number of cards (24) by the number of helpers (3), resulting in each helper receiving 8 cards.

Ms. Taylor's students are giving out cards to each of the 3 class parent helpers equally, and there are a total of 24 cards. To find out how many cards each helper gets, we can divide the total number of cards by the number of helpers.

Step 1: Determine the total number of cards: 24.

Step 2: Determine the number of class parent helpers: 3.

Step 3: Divide the total number of cards by the number of helpers: 24 ÷ 3.

Step 4: Calculate: 24 ÷ 3 = 8.

So, each class parent helper will receive 8 cards.

Answer:

Step-by-step explanation:

From the diagram

b is the length of side adjacent to the angle (x) in the question

a is the length of side opposite to the angle

c is the length of hypotenuse

θ represents the measure of the angle in either degree or radians

From the diagram

Cosecant (csc θ) = c/a

Secant (sec θ) = c/b

Cotangent (cot θ) = b/a

The trigonometric functions in a right triangle are defined as the ratios of the sides: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent.

In a right triangle with sides of lengths a, b, and c (where c is the hypotenuse) and an angle θ, the trigonometric functions are defined as follows:

These functions can easily be remembered using the mnemonic "SOHCAHTOA" - Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, and Tangent equals Opposite over Adjacent.

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Answer:

[tex]P(3\leq X \leq 5)=0.09882[/tex]

Step-by-step explanation:

1) 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".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=6, p=0.2)[/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]

And we want to find this probability:

[tex]P(3 \leq x \leq 5)=P(X=3)+P(X=4)+P(X=5)[/tex]

[tex]P(X=3)=(6C3)(0.2)^3 (1-0.2)^{6-3}=0.08192[/tex]

[tex]P(X=4)=(6C4)(0.2)^4 (1-0.2)^{6-4}=0.01536[/tex]

[tex]P(X=5)=(6C5)(0.2)^5 (1-0.2)^{6-5}=0.001536[/tex]

[tex]P(3 \leq x \leq 5)=0.08192+0.01536+0.001536=0.09882[/tex]

Answer:

The correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.

Step-by-step explanation:

Consider the provided information.

The t-distribution of the Student is a distribution of probability that is used when when the sample size is small and/or when the population variance is unknown to estimate population parameters.

The number of independent observations is equal to the sample size minus one when calculating a mean score or a proportion from a single sample.

Since µ and σ determine the shape of the distribution so we use standard normal distribution if σ is known.

Hence, the correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.

Answer:

p-value 0.0124

Step-by-step explanation:

Hello!

In this experiment, 10 plots were randomly sampled and each plot was divided in a half, one half was planted with the common variety of tomato and the other half was planted with the new hybrid variety. At harvest time it was measured the pounds per plant harvested and the variable difference was established. This study variable is defined as the difference between the pounds per plant of the hybrid variety minus the pound per plant of the common variety. Symbolized: Xd: Xnew - Xcommon.

The purpose of this experiment is to test whether there is a difference between the yield of both species. This is a classic example of a paired test, in which you want to put two dependent samples to test. The easiest way to recognize this type of test, if it is not specified in the problem, is that both variables are measured in the same sampling unit. In this case, the sampling unit is "one plot" that was divided and both verities of tomatoes were planted on it.

This test is also called "paired samples T-test" and as a statistic, you have to use the Student t.

The data given is:

sample: 10 plots

sample mean: Xd[bar]: 1.05ppp

sample standard deviation Sd: 1.23ppp

If the hybrid has a higher yield, that would mean that the difference between them would be positive if the difference is positive, its safe to assume that the population mean of the difference will be positive as well. Symbolically: μd>0

With this, you can state the hypothesis as:

H₀:μd≤0

H₁:μd>0

α: 0.05

t= Xd[bar] - μd ~ [tex]t_{n-1}[/tex]

       Sd/n

t[tex]t_{obs}[/tex]= (1.05 - 0) / (1.23/√10) = 2.69

This is a one-tailed test with critical value [tex]t_{9;0.95} = 1.83[/tex]

the p-value for this test is also one tailed.

p([tex]t_{9}[/tex]≥2.69) = 1-P([tex]t_{9}[/tex]<2.69) = 1 - 0.9876 = 0.0124

The decision is to reject the null hypothesis.

I hope it helps!

Answer:

Confidence interval:  (1760,1956)

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 81

Sample mean =

[tex]\bar{x} = 1858 \text{ kWh}[/tex]

Population standard deviation =

[tex]\sigma = 450 \text{ kilowatt-hours}[/tex]

Confidence Level = 95%

Significance level = 5% = 0.05

Confidence interval:

[tex]\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]

[tex]1858 \pm 1.96(\displaystyle\frac{450}{\sqrt{81}} ) = 1858 \pm 98 = (1760,1956)[/tex]

Answer:

I would assume that the shadow length is proportional to the height of the flagpole. Hence as the building is 5 times higher than the flagpole, the length of the shadow would be 5 times higher, 25x5= 125 ft.

Step-by-step explanation:

The solution is 125 feet

The length of the building's shadow is 125 feet

What is Proportion?

The proportion formula is used to depict if two ratios or fractions are equal. The proportion formula can be given as a: b::c : d = a/b = c/d where a and d are the extreme terms and b and c are the mean terms.

Given data ,

Let the length of the building's shadow be = A

Now , the equation will be

Let the height of the flagpole be = 40 feet

Let the length of the shadow of flagpole be = 25 feet

Let the height of the building be = 200 feet

So , the proportion between the height and length is given by

Height of the flagpole / length of the shadow of flagpole = height of the building / length of the building's shadow

Substituting the values in the equation , we get

40 / 25 = 200 / A

On simplifying the equation , we get

1.6 = 200 / A

Multiply by A on both sides of the equation , we get

1.6A = 200

Divide by 1.6 on both sides of the equation , we get

A = 200/1.6

The value of A = 125 feet

Therefore , the value of A is 125 feet

Hence , the length of the shadow is 125 feet

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Answer:

so (x,y) = (2,3)

Step-by-step explanation:

we have

x +y= 5  ------ equation 1

2x+y= 7-------equation 2

Multiplying equation--1 by 2, we will have equation---3

2x +2y= 10-----equation 3

now subtracting equation 2 from equation 3

2x + 2y= 10

2x + y = 7

     y = 3

adding value of y in equation 1

x + y = 5

x= 5 -y

x= 5 - 3

x =2

so (x,y) = (2,3)

Answer: the correct answer is

5 - x = 7 - 2x

Step-by-step explanation:

Final answer:

To find the added width of the rectangular plot, set up the equation (32 + x)(25 + x) = 2(32)(25) based on the fact that the area is doubled. Solving this equation will give the added width.

Explanation:

To find the added width of the rectangular plot, we need to determine the increase in length and width that will double the area. The original dimensions of the plot are 32 feet long and 25 feet wide. Since equal width is added to each side, the new length and width will be 32 + x and 25 + x, respectively. To find x, we can set up the equation (32 + x)(25 + x) = 2(32)(25) using the fact that the area is doubled. Solving this equation will give us the added width.

Expanding (32 + x)(25 + x), we get 800 + 57x + x^2 = 1600. Rearranging and simplifying, we get x^2 + 57x - 800 = 0. We can then solve this quadratic equation using factoring or the quadratic formula to find the value of x.

The correct equation to use in order to find the added width is (x + 32)(x + 25) = 800.

Final answer:

By setting up a system of equations based on the given information, we find that 200 Basic plans and 80 Standard plans were purchased by the new subscribers.

Explanation:

A cable network offers two types of plans: a Basic plan for $7.31 per month and a Standard plan which costs $3.00 more per month, totaling $10.31. To find out how many Basic and Standard plans were purchased by 280 new subscribers, who altogether paid $2286.80, we can set up a system of equations. Let x represent the number of Basic plans and y represent the number of Standard plans.

The total amount of plans purchased: x + y = 280

The total amount paid: 7.31x + 10.31y = 2286.80

Solving this system of equations:
Step 1: Multiply the first equation by 7.31 to make the x coefficients equal: 7.31x + 7.31y = 2046.80

Step 2: Subtract this equation from the second equation to eliminate x and solve for y: 3y = 240, thus y = 80. Therefore, 80 Standard plans were purchased.

Step 3: Substitute y back into one of the original equations to solve for x: x + 80 = 280, thus x = 200. Therefore, 200 Basic plans were purchased.

So, there were 200 Basic plans and 80 Standard plans purchased by the new subscribers.

Answer:

There was no error

Step-by-step explanation:

There are two types of statistical errors, the type 1 error and the type 2 error. In this case we refute the null hypothesis when the hypothesis is, in fact, false, because the mean process is 9 days instead of 7. Therefore we made no errors.

If the null hypothesis were True, a type 1 error would have ocurred. If the null hypothesis were false and we didnt refute it, then a type 2 error would have ocurred.

Answer:

Step-by-step explanation:

Answer:

  A. ∠A

Step-by-step explanation:

The shortest side is BC. The smallest angle is opposite the shortest side, so is angle A.

∠A is the smallest in measure.

Answer:

[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]  

[tex]p_v =P(z<0.119)=0.547[/tex]  

The p value obtained was a very high value and using the significance level given for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.  

Step-by-step explanation:

1) Data given and notation

n=86 represent the random sample taken

X=59 represent the adults that match the condition you are testing

[tex]\hat p=\frac{59}{86}=0.686[/tex] estimated proportion of adults that match the condition you are testing

[tex]p_o=0.68[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the population proportion is less than 0.68.:  

Null hypothesis:[tex]p=0.68[/tex]  

Alternative hypothesis:[tex]p < 0.68[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided is not given [tex]\alpha[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<0.119)=0.547[/tex]  

So the p value obtained was a very high value and using the significance level given for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.  

Answer:

C) A person's height, recorded in inches

Step-by-step explanation:

Quantitative Variable:

A) The color of an automobile

The color of car is not a quantitative variable as its outcome cannot be measured and expressed in value. It is a categorical variable.

B) A person's zip code

Some variables like zip codes take numerical values. But they are not considered quantitative. They are considered as a categorical variable because average of zip codes have no significance.

C) A person's height, recorded in inches

Height is a qualitative variable because it can be measured and its value is expressed in numbers.

C) A person's height, recorded in inches

A variable can be defined as the characteristics of an object.

A quantitative variable is a variable that usually changes it values as a result of either counting or measuring. Examples are time, weight, height.

A qualitative variable is a variable which do not usually change as a result of counting or measurement. Such as the color of an object, zip code, phone number, license number.

Height is a quantitative variable.

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