In a study of cereal leaf beetle damage to oats, researchers measured the number of beetle larvae per stem in small plots of oats after applying (or not applying) the pesticide Malathion . Researchers applied Malathion to a random sample of 5 plots, and did not apply the pesticide to an independent sample of 12 plots. A noted scientist claims that Malathion will not make any difference in the mean number of larvae per stem. Test her claim at the .05 level of significance. State H0 and H1. a. H0: Mean Malathion -MeanNoMalathion <= 0 H1: Mean Malathion-Mean Malathion > 0 b. H0: MeanMalathion -MeanNoMalathion >= 0 H1: MeanMalathion-MeanNoMalathion < 0 c. H0: MeanMalathion = MeanNoMalathion H1: MeanMalathion ≠ MeanNoMalathion d. H0: MeanMalathion-MeanNoMalathion = 0 H1: MeanMalathion-MeanNoMalathion ≠ 0

The game of chance discussed is a question about probability and expected value in mathematics. To calculate the expected winnings of the game, we use given game information and probabilities. If the probabilities are not given, the question usually assumes a fair spinner, i.e., all outcomes are equally likely.

The subject at hand deals with probability and expected value, which are mathematical concepts typically covered in a high school math curriculum. The game described illustrates these concepts. Each possible outcome of the game (A or D, otherwise lose) corresponds to an event that has a certain probability. These probabilities are all added together to determine the expected value of the game in dollars.

Suppose the probabilities of landing on A and D are p(A) and p(D), and the probability of not landing on either A or D is 1 - p(A) - p(D), then the expected value of the game is: Expected Value = $2 * [p(A)*0.25 + p(D)*9 + (1 - p(A) - p(D))*0] .

To find the expected value, we would need to know the probabilities of landing on each of these segments on the spinner. If these probabilities are not given in the problem, it can be assumed that the spinner is fair (i.e., all outcomes are equally likely). If there are n total segments on the spinner, then p(A) = p(D) = 1/n, and the probability of not landing on A or D would be (n-2)/n. Substitute these probabilities into the expected value equation can give the answer.

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Answer:

The base of the triangle is decreasing at a rate 8 centimeter per minute.      

Step-by-step explanation:

We are given the following in the question:

[tex]\dfrac{dh}{dt} = 2.5\text{ cm per minute}\\\\\dfrac{dA}{dt} = 2\text{ square cm per minute}[/tex]

Instant height = 7.5 cm

Instant area = 96 square centimeters

Area of triangle =

[tex]A=\dfrac{1}{2}\times b\times h[/tex]

where b is the base and h is the height of the triangle.

[tex]96 = \dfrac{1}{2}\times b \times 7.5\\\\b = \dfrac{96\times 2}{7.5} = 25.6[/tex]

Rate of change of area of triangle =

[tex]\dfrac{dA}{dt} = \dfrac{d}{dt}(\dfrac{bh}{2})\\\\\dfrac{dA}{dt} =\dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})[/tex]

Putting values, we get,

[tex]2 = \dfrac{1}{2}(7.5\dfrac{db}{dt}+25.6(2.5))\\\\4 = 7.5\dfrac{db}{dt} + 64\\\\-60 = 7.5\dfrac{db}{dt} \\\\\Rightarrow \dfrac{db}{dt} = \dfrac{-60}{7.5} = -8[/tex]

Thus, the base of the triangle is decreasing at a rate 8 centimeter per minute.

Answer:

a) For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15

b) [tex]P(X=0)=(10C0)(0.15)^0 (1-0.15)^{10-0}=0.1969[/tex]

c) [tex]P(X=3)=(10C3)(0.15)^3 (1-0.15)^{10-3}=0.1298[/tex]

d) [tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)[/tex]

And using the result from part a we got:

[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)= 1-0.1969 =0.8031[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

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

[tex]X \sim Binom(n p)[/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]  

Solution to the problem

Part a

For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15

Part b

For this case we want this probability:

[tex] P(X=0)[/tex]

And replacing we got:

[tex]P(X=0)=(10C0)(0.15)^0 (1-0.15)^{10-0}=0.1969[/tex]

Part c

For this case we want this probability:

[tex] P(X=3)[/tex]

And replacing we got:

[tex]P(X=3)=(10C3)(0.15)^3 (1-0.15)^{10-3}=0.1298[/tex]

Part d

For this cae we want thi probability:

[tex] P(X \geq 1)[/tex]

And we can use the complment rule and we got:

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

And using the result from part a we got:

[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)= 1-0.1969 =0.8031[/tex]

Final answer:

To keep the weekly revenue constant while demand drops, we can set up an equation using the revenue formula. By equating the original revenue with the new revenue, we can find the rate at which the price needs to be raised. Taking the derivative, we can determine the rate of change of the price.

Explanation:

To keep the weekly revenue constant, we need to find the rate at which the price has to be raised to offset the drop in demand. Currently, the price is 40¢ per cup and demand is dropping at a rate of 4 cups per week. Since revenue is equal to price times quantity, we can set up the equation:
Revenue = Price × Quantity.

Initially, we have 80 cups of lemonade sold at 40¢ per cup, resulting in a revenue of $32 (80 x 40¢). As demand drops by 4 cups per week each week, the new quantity sold can be represented by 80 - 4t, where t represents the number of weeks. Let P be the new price per cup that needs to be raised. The new revenue equation can be written as:

Revenue = P(80 - 4t).

To find the value of P, we equate the original revenue ($32) with the new revenue:

$32 = P(80 - 4t).

Simplifying the equation, we get:

32 = 80P - 4Pt.

Moving the terms around, we have:

4Pt = 80P - 32.

Dividing both sides by 4P, we get:

t = (80P - 32)/(4P).

So, the rate at which the price needs to be raised to keep the weekly revenue constant is given by the derivative of t with respect to P. Taking the derivative, we get:

t' = (4(80P - 32) - 4P(80))/(4P)^2.

Simplifying further, we have:

t' = (320P - 128 -  320P)/(4P)^2.

Simplifying again, we get:

t' = -128/(4P)^2.

Thus, the rate of change of t with respect to P is given by -128/(4P)^2. This represents the rate at which the price needs to be raised in order to keep the weekly revenue constant.

Answer: see the pictures attached

Step-by-step explanation:

Final answer:

To model the mosquito population considering both exponential growth and daily predation, a differential equation was formulated and solved, revealing how the population changes over time.

Explanation:

To determine the population of mosquitoes in the area at any time, given that the population doubles each week and predators eat 20,000 mosquitoes per day, we can set up a differential equation. To start, we know the initial population is 200,000 mosquitoes. Given the population increases proportionally, we use the formula P(t) = P_0e^{rt}, where P(t) is the population at time t, P_0 is the initial population, r is the rate of growth, and t represents time in weeks.

To find r, we use the fact that the population doubles each week. So, when t = 1, P(t) = 2P_0, leading to 2P_0 = P_0e^{r(1)}, simplifying to 2 = e^r, which gives r = ln(2).

Including the effect of predators, the amended differential equation becomes dP/dt = rP - 20,000. Substituting r with ln(2) and solving this equation gives us the mosquito population at any time, accounting for both natural growth and predation.

Answer:

a) p = 0.6346

b) 95% confidence interval

Lower limit: 0.5951

Upper limit: 0.6741      

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 572

Number of Santa Fe black-on-whitepots , x = 363

a) proportion of Santa Fe black-on-white potsherds

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{363}{572} = 0.6346[/tex]

b) 95% confidence interval

[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

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

Putting the values, we get:

[tex]0.6346\pm 1.96(\sqrt{\frac{0.6346(1-0.6346)}{572}}) = 0.6346\pm  0.0395\\\\=(0.5951,0.6741)[/tex]

Lower limit: 0.5951

Upper limit: 0.6741

Answer:

51% of US adults favored pepperoni or mushrooms on their cheese pizza.                                                    

Step-by-step explanation:

We are given the following in the question:

Percentage of US adults that favored pepperoni = 43%

[tex]P(P) = 0.43[/tex]

Percentage of US adults that favored mushroom = 14%

[tex]P(M) = 0.14[/tex]

Percentage of US adults that favored both pepperoni and mushroom = 6%

[tex]P(M\cap P) = 0.06[/tex]

We have to evaluate the probability that a random adult favored pepperoni or mushrooms on their cheese pizza.

Thus, we have to evaluate:

[tex]P(M\cup P) = P(M) + P(P) - P(M\cap P)\\P(M\cup P) = 0.43 + 0.14 - 0.06\\P(M\cup P) = 0.51 = 51\%[/tex]

Thus, 51% of US adults favored pepperoni or mushrooms on their cheese pizza.

The solutions to the equation tan(θ) = 1 within the specified range 0 ≤ θ < 2π: θ = 0.7854, 3.9270

Apply the inverse tangent function:

We begin by applying the inverse tangent function (arctan) to both sides of the equation: arctan(tan(θ)) = arctan(1)

Since arctan is the inverse of tangent, they cancel each other out on the left side, leaving us with: θ = arctan(1)

Determine the reference angle:

arctan(1) = π/4, which is the reference angle in the first quadrant where tangent is 1.

Find solutions in other quadrants:

The tangent function has a period of π, meaning it repeats its values every π radians.

Since tangent is also positive in the third quadrant, we add π to the reference angle to find the solution in that quadrant: θ = π/4 + π = 5π/4

Consider the specified range:

We're given the range 0 ≤ θ < 2π. Both π/4 and 5π/4 fall within this range, so they are the valid solutions.

Therefore, the solutions to the equation tan(θ) = 1 within the specified range are θ = 0.7854 (π/4) and θ = 3.9270 (5π/4).

Final answer:

To solve the equation tan(θ) = 1 for θ, we need to use the inverse tangent function. The solution to the equation is θ = π/4.

Explanation:

To solve the equation tan(θ) = 1 for θ, we need to use the inverse trigonometric function. In this case, we will use the inverse tangent function, also known as arctan or atan.

Applying the inverse tangent function to both sides of the equation, we get θ = atan(1).

Using a calculator, we find that atan(1) = π/4. Therefore, the solution to the equation is θ = π/4.

Final answer:

The question involves using Green's Theorem to calculate work done by a specific force field along a triangular path. Green's Theorem links the line integral around a closed curve to a double integral over the region it encloses. However, direct calculation of line integrals may be more applicable for this problem.

Explanation:

The question asks to use Green's Theorem to calculate the work done by a force F(x, y) = x(x + y) i + xy^2 j moving a particle along a specified path. Green's Theorem relates a line integral around a simple closed curve C, to a double integral over the plane region D bounded by C. It's expressed in the form ∑_C (M dx + N dy) = ∫∫_D (∂N/∂x - ∂M/∂y) dA where M and N are components of a vector field.

To use Green's Theorem, we identify M = x(x + y) and N = xy^2. Thus, ∂N/∂x = y^2 and ∂M/∂y = x. The path described (origin to (6,0), to (0,6), and back to origin) encloses a triangle, the area of which is easily integrated over. However, it is crucial to note that direct integration methods or alternative strategies may sometimes be more straightforward for such paths, especially when they do not form a typical closed curve as in standard Green's Theorem applications.

In the context of this problem, the key would be computing the line integrals directly due to the specific nature of the force field and the path involved. The use of Green's Theorem hints at setting up an integration over the area, but with the given vector field and triangular path, the execution would involve careful calculation of respective line integrals or assessing the area directly since the theorem simplifies to a calculation over the region enclosed by the path.

Answer: y = 5x − 11

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

The line passes through (3,4) and (2, -1),

y2 = - 1

y1 = 4

x2 = 2

x1 = 3

Slope,m = (- 1 - 4)/(2 - 3) = - 5/- 1 = 5

To determine the y intercept, we would substitute x = 3, y = 4 and m= 5 into

y = mx + c. It becomes

4 = 5 × 3 + c

4 = 15 + c

c = 4 - 15 = - 11

The equation becomes

y = 5x - 11

Answer:

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 50, p = 0.13[/tex]

So

[tex]\mu = E(X) = np = 50*0.13 = 6.5[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.13*0.87} = 2.38[/tex]

Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

This is 1 subtracted by the pvalue of Z when X = 4. So

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

[tex]Z = \frac{4 - 6.5}{2.38}[/tex]

[tex]Z = -0.84[/tex]

[tex]Z = -0.84[/tex] has a pvalue of 0.2005

1 - 0.2005 = 0.7995

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

Using the normal approximation to the binomial distribution, the probability that at least 5 out of 50 motorists will be uninsured when 13% of motorists are uninsured is approximately 74.22%.

First, we calculate the mean (μ) and the standard deviation (σ) of the number of uninsured motorists in a sample of 50. The mean number of uninsured motorists is μ = np, where n is the sample size and p is the probability of a motorist being uninsured. So, μ = 50 * 0.13 = 6.5.

The standard d eviation is σ = √(np(1-p)). Therefore, σ = √(50 * 0.13 * (1 - 0.13)) ≈ 2.31.

To find the probability of at least 5 uninsured motorists, we want P(X ≥ 5). We need to standardize this to the standard normal distribution to find the z-score. The z-score is z = (X - μ) / σ.

For X = 5, the z-score is z = (5 - 6.5) / 2.31 ≈ -0.65. We use the z-table or a calculator to find the probability corresponding to z = -0.65.

Since we want at least 5 uninsured, we find P(Z ≥ -0.65), which equals 1 - P(Z < -0.65). The value from the z-table for -0.65 is approximately 0.2578. Therefore, the probability we're looking for is 1 - 0.2578 = 0.7422, or 74.22%.

Answer:

Total Cost = Fixed Cost as x --> ∞

Step-by-step explanation:

C'(x) = 4500 x⁻¹•⁹ where x ≥ 1

Marginal Cost = C'(x) = (dC/dx)

C(x) = ∫ (marginal cost) dx

C(x) = ∫ (4500 x⁻¹•⁹)

C(x) = (-5000 x⁻⁰•⁹) + k

where k = constant of integration or in economics term, K = Fixed Cost.

C(x) = [-5000/(x⁰•⁹)] + Fixed Cost

The company wants to make infinitely many units, that is, x --> ∞

C(x --> ∞) = [-5000/(∞⁰•⁹)] + Fixed Cost

(∞⁰•⁹) = ∞

C(x --> ∞) = [-5000/(∞)] + Fixed Cost

But mathematically, any number divide by infinity = 0;

(-5000/∞) = 0

C(x --> ∞) = 0 + Fixed Cost = Fixed Cost.

Total Cost of producing infinite number of units for this cost function is totally the Fixed Cost.

Final answer:

Nickel does not crystallize in a simple cubic structure because the hypothetical density of a simple cubic arrangement is 2.23 g/cm³, significantly less than the actual density of nickel, which is 8.90 g/cm³.

Explanation:

To determine if nickel crystallizes in a simple cubic structure, we can compare the calculated density of a simple cubic arrangement of nickel atoms to the actual density of nickel. Given the edge length of the unit cell for nickel is 0.3524 nm (3.524 x 10⁻⁸ cm) and knowing the properties of a simple cubic lattice, the volume of a single cubic unit cell would be V = a³ = (3.524 x 10⁻⁸ cm)³ = 4.376 x 10⁻²³ cm³. With the mass of a single nickel atom (58.693 g/mol divided by Avogadro's number, 6.022 x 10²³ atoms/mol), if the nickel crystallized in a simple cubic structure, the predicted density would be:

density = mass/volume = (9.746 x 10⁻²³ g) / (4.376 x 10⁻²³ cm³) = 2.23 g/cm³. However, because the actual density of nickel is 8.90 g/cm³, we can conclude that nickel does not form a simple cubic structure.

Answer:

Mean = 1.57

Variance=0.31

Step-by-step explanation:

To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:

1) Both surgeries are successful (X=2).

[tex]P(e_1)=0.90*0.67=0.603[/tex]

2) Left knee unsuccessful and right knee successful (X=1).

[tex]P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067[/tex]

3) Right knee unsuccessful and left knee successful (X=1).

[tex]P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297[/tex]

4) Both surgeries are unsuccessful (X=0).

[tex]P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033[/tex]

Then, the mean can be calculated as the expected value:

[tex]M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57[/tex]

The variance can be calculated as:

[tex]V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111[/tex]

The mean and variance of the number of successful surgeries for both knees combined are:

Mean: [tex]\({1.57}\)[/tex]

Variance: [tex]\({0.3111}\)[/tex]

The mean and variance of the number of successful surgeries for the given meniscal tears can be calculated using the information provided about the success rates.

For a meniscal tear with a rim width of less than 3 mm, the success rate is more than 90%. For simplicity, let's assume the success rate is exactly 90% (since we don't have the exact number above 90%). For a tear of 3-6 mm, the success rate is 67%.

Let's denote the success of a surgery as a random variable [tex]\( X \)[/tex], which takes the value 1 if the surgery is successful and 0 if it is not. The probability of success [tex]\( P(X = 1) \)[/tex] is the success rate, and the probability of failure [tex]\( P(X = 0) \)[/tex] is [tex]\( 1 - P(X = 1) \)[/tex].

For the left knee (tear less than 3 mm):

- [tex]\( P(X = 1) = 0.90 \)[/tex] (success rate)

- [tex]\( P(X = 0) = 1 - 0.90 = 0.10 \)[/tex] (failure rate)

For the right knee (tear 3-6 mm):

- [tex]\( P(X = 1) = 0.67 \)[/tex] (success rate)

- [tex]\( P(X = 0) = 1 - 0.67 = 0.33 \)[/tex] (failure rate)

The mean (expected value) of the number of successful surgeries for each knee is calculated as follows:

For the left knee:

[tex]\[ E(X) = \sum_{i=0}^{1} x_i \cdot P(X = x_i) = 1 \cdot 0.90 + 0 \cdot 0.10 = 0.90 \][/tex]

For the right knee:

[tex]\[ E(X) = \sum_{i=0}^{1} x_i \cdot P(X = x_i) = 1 \cdot 0.67 + 0 \cdot 0.33 = 0.67 \][/tex]

The variance of the number of successful surgeries for each knee is calculated using the formula for the variance of a binary random variable:

For the left knee:

[tex]\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]

[tex]\[ E(X^2) = \sum_{i=0}^{1} x_i^2 \cdot P(X = x_i) = 1^2 \cdot 0.90 + 0^2 \cdot 0.10 = 0.90 \][/tex]

[tex]\[ \text{Var}(X) = 0.90 - (0.90)^2 = 0.90 - 0.81 = 0.09 \][/tex]

For the right knee:

[tex]\[ E(X^2) = \sum_{i=0}^{1} x_i^2 \cdot P(X = x_i) = 1^2 \cdot 0.67 + 0^2 \cdot 0.33 = 0.67 \][/tex]

[tex]\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]

[tex]\[ \text{Var}(X) = 0.67 - (0.67)^2 = 0.67 - 0.4489 = 0.2211 \][/tex]

Now, assuming the surgeries on the two knees are independent events, the mean and variance for both knees combined can be calculated as follows:

Mean for both knees:

[tex]\[ E(X_{\text{left}} + X_{\text{right}}) = E(X_{\text{left}}) + E(X_{\text{right}}) = 0.90 + 0.67 = 1.57 \][/tex]

Variance for both knees:

Since the surgeries are independent, the variance of the sum is the sum of the variances:

[tex]\[ \text{Var}(X_{\text{left}} + X_{\text{right}}) = \text{Var}(X_{\text{left}}) + \text{Var}(X_{\text{right}}) = 0.09 + 0.2211 = 0.3111 \][/tex]

Therefore, the mean and variance of the number of successful surgeries for both knees combined are:

Mean: [tex]\({1.57}\)[/tex]

Variance: [tex]\({0.3111}\)[/tex]

Answer:

a. Predicted Amount = $109.46

b. Confidence Interval = (94.84,124.08)

c. Interval = (110.6883,188.8517)

Step-by-step explanation:

Given

ŷ = 17.49 + 1.0334x.

SSE = 1541.4

a.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46 --- Approximated

Predicted Amount = $109.46

b.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46

First we calculate the standard deviation

variance = SSE/(n-2)

v = 1541.4/(9-2)

v = 1541.4/7

v = 220.2

s = √v

s = √220.2

s = 14.839

Then we calculate mean(x) and ∑(x - (mean(x))²

X --- Y -- Mean(x) --- ∑(x - (mean(x))²

148 -- 161 -- 43-- 1849

96 || 105|| -9 || 81

91 ||101 || -14 || 196

110 || 142 || 5 || 25

90 || 100 || -15 || 225

102 || ||120 ||-3|| 9

136 || 167 ||31 ||961

90 || 140 ||-15 ||225

82 || 98 ||-23 || 529

Sum 945 || 1134|| 0 ||4100

Mean (x) = 945/9 = 105

∑(x - (mean(x))² = 4100

α = 1 - 95% = 5%

α/2 = 2.5% = 0.025

tα,df = n − 2 = t0.025,7 =2.365

Confidence interval = 109.46 ± 2.365 * 14.839 √((1/9)+ (89-105)²/4100

Confidence Interval = (109.46 ± 14.62)

Confidence Interval = (94.84,124.08)

c.

ŷ = 17.49 + 1.0334(128)

ŷ = 149.7652

ŷ = 149.77

Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100

Interval = 149.77 ± 39.0817

Interval = (110.6883,188.8517)

Given the regression equation ŷ = 17.49 + 1.0334x, we can predict the amount spent on entertainment in cities based on their daily room rate. For instance, a city with a daily room rate of $89 is estimated to spend about $109.67 on entertainment. However, we don't have enough information to calculate the 95% confidence interval or the 95% prediction interval.

To solve these questions, we use the provided regression equation, which is ŷ = 17.49 + 1.0334x. The variable 'x' represents the daily room rate, and 'ŷ' represents the predicted amount spent on entertainment.

(a) To predict the amount spent on entertainment for a city that has a daily room rate of $89, substitute x with 89 in the equation: ŷ = 17.49 + 1.0334 * 89. The computed prediction is $109.67.

(b) To develop a 95% confidence interval for the mean amount spent on entertainment for all cities with a daily room rate of $89, we would need additional statistical data such as the standard error or the number of data points. There isn't sufficient information in the question to accurately compute this.

(c) To find the 95% prediction interval for the amount spent on entertainment in Chicago with an average room rate of $128, we would also need additional statistical data like the standard error, degrees of freedom, or the number of observations. Again, the question does not provide sufficient details to calculate this.

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Question

A salesperson obtained a systematic sample of size 25 from a list of 500 clients. To do​ so, he randomly selected a number from 1 to 20, obtaining the number 13. He included in the sample the 13th client on the list and every 20th client thereafter. List the numbers that correspond to the 25 clients selected.

Answer:

13, 33, 53, ...... 393

Step-by-step explanation:

The Question is making reference to arithmetic progression

Where the first term is 13

And the difference between each interval is 20 (i.e number of clients)

Using the following formula, well obtained the client in any position

Tn = a + (n - 1) d

Where Tn = nth term

a = T1 = 1st term = 13

n = number of terms = 1 to 25

d = common difference = 20

Calculating T2

T2 = 13 + (2 - 1) * 20

T2 = 13 + 20

T2 = 33

Calculating T3

T3 = 13 + (3 - 1) * 20

T3 = 53

.......

Calculating the 20th term

T3 = 13 + (20 - 1) * 20

T20 = 393

Answer:

Step-by-step explanation:

Given data:

Mass of the one arm of the skater, m = (1/16) x 64 = 4 kg

Rest mass of the skater in the form of cylinder, M = (7 / 8) x 64 kg = 56 kg

Radius of the cylinder, R = 20 cm = 0.20 m

The parallel axis theorem:

Answer:

95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (60.77% , 65.23%)

Step-by-step explanation:

Among 1800 NAU students, 1134 students love their class. We have to find the 95% confidence interval of students who love their class.

We will use the concept of confidence interval of population proportion for this problem.

The proportion of students who love the class = p = [tex]\frac{1134}{1800}=0.63[/tex]

Proportion of students who do not love the class = q = 1 - p = 1 - 0.63 = 0.37

Total number of students in the sample = n = 1800

Confidence Level = 95%

The z values associated with this confidence level(as seen from z table) = 1.96

The formula to calculate the confidence interval for population proportion is:

[tex](p-z\times\sqrt{\frac{p \times q}{n}},p+z\times\sqrt{\frac{p \times q}{n}})[/tex]

Using the values in this expression gives:

[tex](0.63-1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}}, 0.63+1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}})\\\\ =(0.6077,0.6523)[/tex]

Thus, 95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (0.6077 ,0.6523) or (60.77% , 65.23%

Answer: the American car gets better gas mileage.

Step-by-step explanation:

American car travels 32 miles on 1 gallon of gas. This means that its gas mileage is 32 miles per gallon.

European car travels 12.7 km on 1 L of gas. We would convert 12.7 km to miles.

1 kilometer = 0.621 mile

12.7 kilometer = 12.7 × 0.621 = 7.8867 miles

We would also convert 1 L to gallons.

1 liter = 0.264 us gallons

Therefore, the gas mileage of the European car in miles per gallon would be

7.8867/0.264 = 29.87 miles per gallon.

Therefore, since 32 miles per gallon is greater than 29.87 miles per gallon, it means that the American car gets better gas mileage.

Answer:

b/a = c/b

if a = b, then b = c

Answer: the second one (b/a = c/b) and the last one (if a = b then b = c) are the only ones that are true

Step-by-step explanation:

Answer:

[tex]P(X = 0) = 0.42[/tex]

[tex]P(X = 1) = 0.46[/tex]

[tex]P(X = 2) = 0.12[/tex]

Step-by-step explanation:

We have these following probabilities:

40% probability that the first file is corrupted. So 60% probability that the first file is not corrupted.

30% probability that the second file is corrupted. So 70% probability that the second file is not corrupted.

Probability mass function

Probability of each outcome(0, 1 and 2 files corrupted).

No files corrupted:

60% probability that the first file is not corrupted.

70% probability that the second file is not corrupted.

So

[tex]P(X = 0) = 0.6*0.7 = 0.42[/tex]

One file corrupted:

First one corrupted, second no.

40% probability that the first file is corrupted.

70% probability that the second file is not corrupted.

First one ok, second one corrupted.

60% probability that the first file is not corrupted.

30% probability that the second file is corrupted.

[tex]P(X = 1) = 0.4*0.7 + 0.6*0.3 = 0.46[/tex]

Two files corrupted:

40% probability that the first file is corrupted.

30% probability that the second file is corrupted.

[tex]P(X = 2) = 0.4*0.3 = 0.12[/tex]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/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 = 300, p = 0.02[/tex]

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

[tex]E(X) = np = 300*0.02 = 6[/tex]

Standard deviation

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42[/tex]

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040[/tex]

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023[/tex]

[tex]P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143[/tex]

[tex]P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436[/tex]

[tex]P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883[/tex]

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485[/tex]

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

From c), we have that [tex]P(X = 0) = 0.0023[/tex]. So

[tex]0.0023 + P(X \geq 1) = 1[/tex]

[tex]P(X \geq 1) = 0.9977[/tex]

99.77% probability that we find at least 1 of the seniors are married

In this problem, the mean and standard deviation of a binomial distribution, with probability of success 0.02 and sample size 300, are found. Subsequently, the probabilities that 8, less than 4, and at least 1 of the seniors are married are computed using the binomial formula.

This problem deals with the Binomial distributions in statistics. Since we know the proportion of high school seniors who are married is 0.02, and the sample size is 300, we can use these values to calculate the mean and the standard deviation.

a.) The mean (mean = np) of the sample count X who are married is 0.02*300=6, and the standard deviation would be sqrt(n*p*(1-p)) = sqrt(300*0.02*0.98) = √5.88≈2.43.

b.) The probability that, in our sample of 300, we find that 8 of the seniors are married is given by the binomial formula P(X=k) = C(n,k)(p^k)(1-p)^(n-k). Plugging n=300, k=8, p=0.02 into the formula, we get the desired probability.

c.) The probability that we find less than 4 of the seniors are married is sum of P(X=k) from k=0 to 3. This could be computed using the aforementioned binomial formula. Remember, you're summing the probabilities for each k.

d.) The probability that we find at least 1 of the seniors are married can be found by subtracting the probability that none of the seniors are married from 1 (i.e., P(X >=1) = 1 - P(X=0)).

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Answer:

D. 0.0384

Step-by-step explanation:

For each loan, there are only two possible outcomes. Either the client makes timely payments, or he does not. The probability of a client making a timely payment is independent from other clients. So we use the binomial probability distribution to solve this question.

However, our sample is big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 300, p = 0.04[/tex]

So

[tex]\mu = E(X) = np = 300*0.04 = 12[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.04*0.96} = 3.39[/tex]

What is the probability that over 6% will not make timely payments?

This is 1 subtracted by the pvalue of Z when X = 0.06*300 = 18. So

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

[tex]Z = \frac{18 - 12}{3.39}[/tex]

[tex]Z = 1.77[/tex]

[tex]Z = 1.77[/tex] has a pvalue of 0.9616

1 - 0.9616 = 0.0384

So the correct answer is:

D. 0.0384

Answer:

4/8 more than 3/ but less than 5/8

Answer: I modeled 4/8 because it is greater than 3 less than 5 2 equivalent fractions are 8/16 12/24

Step-by-step explanation:

Answer:

(x,y)=(1,-4)

Step-by-step explanation:

y=−7x+3

y=−x−3 ​

(y=) −7x+3=−x−3 ​

-7x+x=-3-3

-6x=-6

x=-6/(-6)

x=1

y=-7*1+3=-7+3=-4

(x,y)=(1,-4)

Answer:

[tex](x,y)= (1,-4)\\[/tex]

Step-by-step explanation:

We will solve it using the substitution method

Using Substitution method

Let [tex]y = -7x + 3[/tex] be equation 1 and [tex]y = -x - 3[/tex] be equation 2

putting value of y from equation 1 in equation 2 and further simplifying:

we get

[tex]-7x +3 = -x - 3\\-7x + x = -3 -3\\-6x =-6\\\\6x=6x\\x= 1[/tex]

Now put value of x i.e. [tex]x=1[/tex] in equation 1 and by further simplifying

[tex]y = -7x + 3\\y= -7(1) +3\\y= -7+3\\y=-4[/tex]

So the solution to the system is written as\[tex](x,y)= (1,-4)[/tex]

Answer:

[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]

And we can find this probability using the normal standard table or excel:

[tex]P(z<-1.25)=0.106[/tex]

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the number of days before cartridge runs out of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(75,6)[/tex]  

Where [tex]\mu=75[/tex] and [tex]\sigma=6[/tex]

We are interested on this probability

[tex]P(X<67.5)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]

And we can find this probability using the normal standard table or excel:

[tex]P(z<-1.25)=0.106[/tex]

Answer:

64.69e221

Step-by-step explanation:

When choosing, the combination formula for selection is used. That is when selecting "r" number of items from a possible "n" items, then the number of ways is denoted as:

nCr = n! / (n-r)! × r!

Since 66 string quartet have to be chosen and the 3 genres must be equally represented in the string quartet, then we must have 22 number of each genre in it.

Number of ways to select 22 mendelssohn from possible 66 = 66C22 = 1.82 × 10^17

Number of ways to select 22 Beethoven from possible 1616 = 1616C22 = 2.97 × 10^49

Number of ways to select 22 Haydn from possible 6868 = 6868C22 = 2.2 × 10^63

Total number of ways to arrange these 66 schedules = 66! = 5.44 × 10^92

Number of possible schedule = [1.82 * 10^17] * [2.97*10^49] * [2.2*10^63] * [5.44*10^92]

=64.69 ×10^221. ≈64.69e221

The current volume of the water is 261.66 square inches.

Solution:

The container is in cone shape.

Radius of the container = 5 inch

Height of the container = 20 inch

Volume of the container = [tex]\frac{1}{3} \pi r^2 h[/tex]

                                        [tex]$=\frac{1}{3}\times 3.14 \times 5^2 \times 20[/tex]

Volume of the container = 523.33 square inch

Current volume of the water = Half of the volume of container

                                               [tex]$=\frac{1}{2}\times523.33[/tex]

                                               = 261.66 square inch

The current volume of the water is 261.66 square inches.

Answer:

We have 252 different schedules.

Step-by-step explanation:

We know that as  a freshman, suppose you had to take two of four lab science courses, one of two literature courses, two of three math courses, and one of seven physical education courses.

So from 4 lab science courses we choose 2:

[tex]C_2^4=\frac{4!}{2!(4-2)!}=6[/tex]

So from 2 literature courses we choose 1:

[tex]C_1^2=\frac{2!}{1!(2-1)!}=2[/tex]

So from 3 math courses we choose 2:

[tex]C_2^3=\frac{3!}{2!(3-2)!}=3\\[/tex]

So from 7 physical education courses we choose 1:

[tex]C_1^7=\frac{7!}{1!(7-1)!}=7[/tex]

We get: 6 · 2 · 3 · 7 = 252

We have 252 different schedules.