A survey among US adults of their favorite toppings on a cheese pizza reported that 43% favored pepperoni, 14% favored mushrooms, and 6% favored both pepperoni and mushrooms. What is the probability that a random adult favored pepperoni or mushrooms on their cheese pizza? Provide your answer as a whole number in the box below, i.e., .32 is 32% so you would enter 32. Round as needed.

Answer:

The z statistic is 0.23.

Step-by-step explanation:

Test statistic (z) = (sample mean - population mean) ÷ sd/√n

sample mean = 104 mg

population mean (mu) = 100 mg

sd = 55 mg

n = 10

z = (104 - 100) ÷ 55/√10 = 4 ÷ 17.393 = 0.23

Final answer:

The z-statistic for the given hypothesis test is calculated based on the sample mean, population standard deviation, and sample size.

Explanation:

The z-statistic for this hypothesis test can be calculated using the sample mean, population standard deviation, and sample size.

Given that the sample mean is 104 mg, population standard deviation is 55 mg, and sample size is 10 pills, the z-statistic is calculated as (104 - 100) / (55 / sqrt(10)), which equals 0.8.

Therefore, the correct z-statistic for this scenario is 0.80.

Answer: $2187 was earned in ticket sales

Step-by-step explanation:

At the play, the total number of guests seated on the main floor is 211. If tickets for the main floor

cost $7, it means that the total cost of 211 ticket s is

211 × 7 = $1477

the total number of guests that were seated in the balcony is 142. If tickets for the main floor

cost $5, it means that the total cost of 142 tickets is

5 × 142 = $710

Therefore, the total amount of money earned in ticket sales is

1477 + 710 = $2187

The total amount earned in ticket sales is $2,187.

To determine the total amount earned in ticket sales at the play, we can use the following information:

- Number of guests on the main floor: 211

- Ticket price for the main floor: $7

- Number of guests in the balcony: 142

- Ticket price for the balcony: $5

We can calculate the earnings from each section separately and then sum them up.

Step-by-Step Calculation:

1. Calculate earnings from the main floor:

[tex]\[\text{Earnings from main floor} = \text{Number of main floor guests} \times \text{Ticket price for main floor}\][/tex]

[tex]\[\text{Earnings from main floor} = 211 \times 7 = 1477\][/tex]

2. Calculate earnings from the balcony:

[tex]\[\text{Earnings from balcony} = \text{Number of balcony guests} \times \text{Ticket price for balcony}\][/tex]

[tex]\[\text{Earnings from balcony} = 142 \times 5 = 710\][/tex]

3. Calculate total earnings:

[tex]\[\text{Total earnings} = \text{Earnings from main floor} + \text{Earnings from balcony}\][/tex]

[tex]\[\text{Total earnings} = 1477 + 710 = 2187\][/tex]

When [tex]x=3[/tex], we get the parabola

[tex]z=-4-2y^2[/tex]

We can parameterize this parabola by

[tex]\vec r(t)=(3,t,-4-2t^2)[/tex]

Then the tangent vector to this parabola is

[tex]\vec T(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=(0,1,-4t)[/tex]

We get the point (3, 2, -12) when [tex]t=2[/tex], for which the tangent vector is

[tex]\vec T(2)=(0,1,-8)[/tex]

Then the line tangent to the parabola at [tex]t=2[/tex] passing through the point (3, 2, -12) has vector equation

[tex]\ell(t)=(3,2,-12)+t(0,1,-8)=(3,2+t,-12-8t)[/tex]

which in parametric form is

[tex]\begin{cases}x(t)=3\\y(t)=2+t\\z(t)=-12-8t\end{cases}[/tex]

for [tex]t\in\Bbb R[/tex].

The parametric equation of the tangent line is [tex]L(t)=(3,2+t,-12-8t)[/tex]

The equation of Paraboloid is,

                 [tex]z =8-x-x^{2} -2y^{2}[/tex]

Equation of parabola when [tex]x = 3[/tex] is,

       [tex]z=8-3-3^{2} -2y^{2} \\\\z=-4-2y^{2}[/tex]

The parametric equation of parabola will be,

     [tex]r(t)=(3,t,-4-2t^{2} )[/tex]

Now, we have to find Tangent vector to this parabola is,

    [tex]T(t)=\frac{dr(t)}{dt}=(0,1,-4t)[/tex]

We get, the point [tex](3, 2, -12)[/tex] when [tex]t=2[/tex]

The tangent vector will be,

 [tex]T(2)=(0,1,-8)[/tex]

So that, the tangent line to this parabola at the point (3, 2, −12) will be,

     [tex]L(t)=(3,2,-12)+t(0,1,-8)\\\\L(t)=(3,2+t,-12-8t)[/tex]

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Answer:

4.45 ccs will remain after 5 hours.

Step-by-step explanation:

The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.

(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)

(dQ/dt) = -kQ

(dQ/Q) = -kdt

 ∫ (dQ/Q) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.

We obtain

In (Q/Q₀) = -kt

(Q/Q₀) = e⁻ᵏᵗ

Q(t) = Q₀ e⁻ᵏᵗ

the initial injection was 8 ccs and 5 ccs remain after 4 hours

Q₀ = 8 ccs,

At t = 4 hours, Q = 5 ccs

5 = 8 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.625

-kt = In (0.625) = -0.47

-4k = 0.47

k = 0.1175 /hour

Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ

At t = 5 hours, Q = ?

Q = 8 e⁻⁰•¹¹⁷⁵ᵗ

0.1175 × 5 = 0.5875

Q = 8 e(^-0.5875)

Q = 4.45 ccs

Hope this Helps!!!

Answer:

22.66% probability that the mean weight will exceed 3.81 ounces

Step-by-step explanation:

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

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 3.78, \sigma = 0.14, n = 12, s = \frac{0.14}{\sqrt{12}} = 0.04[/tex]

What is the probability that the mean weight will exceed 3.81 ounces

This probability is 1 subtracted by the pvalue of Z when X = 3.81. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{3.81 - 3.78}{0.04}[/tex]

[tex]Z = 0.75[/tex]

[tex]Z = 0.75[/tex] has a pvalue of 0.7734

1 - 0.7734 = 0.2266

22.66% probability that the mean weight will exceed 3.81 ounces

Answer:

Correct answer is 0.2290

Step-by-step explanation:

Answer:

For k = 6 or k = -6, the equation will have exactly one solution.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

[tex]ax^{2} + bx + c, a\neq0[/tex].

This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:

[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]

[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]

[tex]\bigtriangleup = b^{2} - 4ac[/tex]

If [tex]\bigtriangleup = 0[/tex], the equation has only one solution.

In this problem, we have that:

[tex]3x^{2} + kx + 3 = 0[/tex]

So

[tex]a = 3, b = k, c = 3[/tex]

[tex]\bigtriangleup = b^{2} - 4ac[/tex]

[tex]\bigtriangleup = k^{2} - 4*3*3[/tex]

[tex]\bigtriangleup = k^{2} - 36[/tex]

We will only have one solution if [tex]\bigtriangleup = 0[/tex]. So

[tex]\bigtriangleup = 0[/tex]

[tex]k^{2} - 36 = 0[/tex]

[tex]k^{2} = 36[/tex]

[tex]k = \pm \sqrt{36}[/tex]

[tex]k = \pm 6[/tex]

For k = 6 or k = -6, the equation will have exactly one solution.

Answer:

The standard error for the mean finish time of 30 randomly selected participants is 0.055 hours

Step-by-step explanation:

Standard error = sample standard deviation ÷ sqrt (sample size)

sample standard deviation = 0.30 hours

sample size = 30

standard error = 0.30 ÷ sqrt(30) = 0.30 ÷ 5.477 = 0.055 hours

a. By the law of total probability,

[tex]P(E)=P(A_1\cap E)+P(A_2\cap E)+P(A_3\cap E)[/tex]

and using the definition of conditional probability we can expand the probabilities of intersection as

[tex]P(E)=P(E\mid A_1)P(A_1)+P(E\mid A_2)P(A_2)+P(E\mid A_3)P(A_3)[/tex]

[tex]P(E)=0.1\cdot0.3+0.6\cdot0.5+0.8\cdot0.2=0.49[/tex]

b. Using Bayes' theorem (or just the definition of conditional probability), we have

[tex]P(A_1\mid E)=\dfrac{P(A_1\cap E)}{P(E)}=\dfrac{P(E\mid A_1)P(A_1)}{P(E)}[/tex]

[tex]P(A_1\mid E)=\dfrac{0.1\cdot0.3}{0.49}\approx0.0612[/tex]

c. Same reasoning as in (b):

[tex]P(A_2\mid E)=\dfrac{P(E\mid A_2)P(A_2)}{P(E)}\approx0.612[/tex]

d. Same as before:

[tex]P(A_3\mid E)=\dfrac{P(E\mid A_3)P(A_3)}{P(E)}\approx0.327[/tex]

(Notice how the probabilities conditioned on [tex]E[/tex] add up to 1)

Answer:

0.1577 or 15.77%

Step-by-step explanation:

The Poisson probability model follows the given relationship:

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

For a Poisson model with a parameter λ = 3.85 Punch Baggies/hour, the probability of X=2 successes (exactly 2 Punch Baggies in an hour) is given by:

[tex]P(X=2) =\frac{3.85^2*e^{-3.85}}{2!}\\P(X=2) =0.1577=15.77\%[/tex]

The probability is 0.1577 or 15.77%.

Multiple linear regression can be said to be a statistical technique that uses several explanatory variables to predict the outcome of a response variable.

The Formula for Multiple Linear Regression Is

yi= β0+β1xi1+β2xi2+...+βpxip+ϵ

​

where, for i=n observations

yi​= Dependent variable

xi​= Expanatory variable

β0​= y - intercept (constant term)

βp​= Slope coefficients for each explanatory variable

ϵ= The model’s error term (also known as the residuals)​

From the question written above,

yi= weight gain which is dependent on the explanatory variables which are hemoglobin change, tap water consumption, and age. i.e. xi

Model assumptions

1. There is a linear relationship between the dependent variables and the independent variables.

2. The independent variables are not too highly correlated with each other.

3. yi observations are selected independently and randomly from the population.

4. Residuals should be normally distributed with a mean of 0 and variance σ.

(Investopedia, 2019)

Answer:

Sew attachment for xomplete solution and answer

Step-by-step explanation:

Given that;

Suppose that f is a continuous function and that −24 ≤ f 00(x) ≤ 3 for 0 ≤ x ≤ 1. If the Midpoint Rule of order n, namely Mn, is used to approximate R 1 0 f(x) dx, how large must n be to guarantee that the absolute error in using Mn ≈ R 1 0 f(x) dx is less than 1/100

See attachment

The correct statement is that the probability of both the students who have volunteered for community service the absolute tolerance will be 0.1056.

The calculation of the probability of both the students who have volunteered for the community service getting selected is shown by doing multiple calculations as under.

Hence, correct statement is that the probability of both the students who have volunteered for community service the absolute tolerance will be 0.1056.

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Answer:

A) P(A⋂B) = 0.35

B) P(A⋃B)= 0.53

C) P(A⋂B′) = 0.08

D) P(A|B) = 0.778

Step-by-step explanation:

We know the following from the question:

- Let Proportion of Americans who expect to save more money next year than they saved last year be

P(A) and its = 0.43

-Let proportion who plan to reduce debt next year be P(B) and it's =0.81

A) probability that this person expects to save more money next year and plans to reduce debt next year which is; P(A⋂B) = 0.43 x 0.81 = 0.348 approximately 0.35

B) probability that this person expects to save more money next year or plans to reduce debt next year which is;

P(A⋃B)= P(A) + P(B) − P(A⋂B)

So, P(A⋃B)= 0.43 + 0.45 − 0.35 = 0.53

C). Probability that this person expects to save more money next year and does not plan to reduce debt next year which is;

P(A⋂B′) = P(A) − P(A⋂B)

P(A⋂B′) =0.43 − 0.35 = 0.08

D) Probability that this person does not expect to save more money given that he/she does plan to reduce debt next year which is;

P(A|B) = [P(A⋂B)] / P(B)

So P(A|B) =0.35/0.45 = 0.778

The probabilities for the given scenarios are as follows: a. 34.93%, b. 53.07%, c. 8.07%, and d. 126.67%.

a. To find the probability that a person expects to save more money next year and plans to reduce debt next year, we need to multiply the probabilities of both events occurring. The probability that a person expects to save more money next year is 43%, and of those who expect to save more money next year, 81% plan to reduce debt. Therefore, the probability is 0.43  imes 0.81 = 0.3493, or 34.93%.

b. To find the probability that a person expects to save more money next year or plans to reduce debt next year, we can add the probabilities of both events occurring and subtract the probability of both events occurring at the same time (found in part a). The probability of expecting to save more money next year is 43%, and the probability of planning to reduce debt next year is 45%. Therefore, the probability is 0.43 + 0.45 - 0.3493 = 0.5307, or 53.07%.

c. To find the probability that a person expects to save more money next year and does not plan to reduce debt next year, we subtract the probability from part a from the probability of expecting to save more money next year. The probability of expecting to save more money next year is 43%, and the probability of both expecting to save more money and planning to reduce debt is 34.93%. Therefore, the probability is 0.43 - 0.3493 = 0.0807, or 8.07%.

d. To find the probability that a person does not expect to save more money given that he/she does plan to reduce debt next year, we need to divide the probability of not expecting to save more money by the probability of planning to reduce debt next year. The probability of not expecting to save more money is 1 - 0.43 = 0.57, and the probability of planning to reduce debt next year is 45%. Therefore, the probability is 0.57 / 0.45 = 1.2667, or 126.67%.

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Answer:

d. -1.9; not unusual

Step-by-step explanation:

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.

In this problem, we have that;

[tex]X = 50, \mu = 69, \sigma = 10[/tex].

So

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

[tex]Z = \frac{50 - 69}{10}[/tex]

[tex]Z = -1.9[/tex]

A z-score of -1.9 is higher than -2 and lower than 2, so it is not unusual.

So the correct answer is:

d. -1.9; not unusual

a. Null hypothesis (H0): Mean cost for the new production method is $215 per hour.

Alternative hypothesis (H1): Mean cost for the new production method is less than $215 per hour.

b. Type I error: Rejecting the null hypothesis when it is true; consequence: Implementing the new method incorrectly and incurring unnecessary costs.

c. Type II error: Failing to reject the null hypothesis when it is false; consequence: Missing the opportunity to adopt a cost-effective production method and continuing with higher operating costs.

We have,

a. The appropriate null and alternative hypotheses would be:

Null Hypothesis (H0): The mean cost for the new production method is equal to or greater than $215 per hour.

Alternative Hypothesis (H1): The mean cost for the new production method is less than $215 per hour.

b. The Type I error in this situation would be rejecting the null hypothesis when it is actually true.

In other words, it would mean concluding that the new production method reduces the mean operating cost per hour when it actually doesn't.

The consequence of making this error is that resources, time, and effort might be invested in implementing the new method based on incorrect information, potentially leading to unnecessary costs and inefficiencies.

c. The Type II error in this situation would be failing to reject the null hypothesis when it is actually false.

In other words, it would mean failing to conclude that the new production method reduces the mean operating cost per hour when it actually does.

The consequence of making this error is that the opportunity to adopt a more cost-effective production method would be missed, potentially leading to continued higher operating costs and missed efficiency gains.

Thus,

a. Null hypothesis (H0): Mean cost for the new production method is $215 per hour.

Alternative hypothesis (H1): Mean cost for the new production method is less than $215 per hour.

b. Type I error: Rejecting the null hypothesis when it is true; consequence: Implementing the new method incorrectly and incurring unnecessary costs.

c. Type II error: Failing to reject the null hypothesis when it is false; consequence: Missing the opportunity to adopt a cost-effective production method and continuing with higher operating costs.

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The null and alternative hypotheses are set as μ ≥ 215 and μ < 215 respectively. A Type I error, rejecting the null hypothesis when it's true, could lead to unnecessary changes. A Type II error, failing to reject the null hypothesis when it's false, could result in missed cost savings.

a. In this situation, the null hypothesis (H0) would state that the mean cost of the new production method is equal to or greater than 215. We can denote this as: H0: μ ≥ 215. The alternative hypothesis (Ha) posits that the mean cost of the new production method is less than 215, denoted as: Ha: μ < 215.

b. A Type I error would occur if we incorrectly reject the null hypothesis, concluding that the new method reduces the cost when, in fact, it does not. The consequence could be implementing a production method that doesn't actually provide the desired savings, which could lead to unnecessary changes and associated costs.

c. A Type II error would happen if we fail to reject the null hypothesis when it's actually false, meaning the new method does reduce the cost but we fail to recognize it. The consequence of this error could be missing out on an opportunity to reduce operating costs, potentially leading to higher than necessary expenses.

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Answer:

(a) P(x=0) = 0.1422

(b) P(x=1) = 0.3012

(c) P(x≥2) = 0.5566

Step-by-step explanation:

The probability that x drivers will have an invalid license follows a binomial distributions, so it is calculated as:

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Where n is equal to the 12 drivers stopped and p is the probability that a driver stopped for speeding have invalid licenses. Then, replacing values, we get:

[tex]P(x)=\frac{12!}{x!(12-x)!}*0.15^{x}*(1-0.15)^{12-x}[/tex]

Now, the probability that none will have an invalid licenses is calculated as:

[tex]P(x=0)=\frac{12!}{0!(12-0)!}*0.15^{0}*(1-0.15)^{12-0}\\P(x=0)=0.1422[/tex]

At the same way, the probability that exactly one will have an invalid license  is calculated as:

[tex]P(x=1)=\frac{12!}{1!(12-1)!}*0.15^{1}*(1-0.15)^{12-1}\\P(x=1)=0.3012[/tex]

Finally, the probability that at least 2 will have invalid licenses is calculated as:

[tex]P(x\geq 2)=1-P(x<2)\\P(x\geq 2)=1-(P(x=0)+P(x=1))\\P(x\geq 2)=1-(0.1422+0.3012)\\P(x\geq 2)=1-0.4434\\P(x\geq 2)=0.5566[/tex]

Answer:

a) [tex] P(X=0) = (12C0) (0.15)^0 (1-0.15)^{12-0} =0.1422[/tex]

b) [tex] P(X=1) = (12C1) (0.15)^1 (1-0.15)^{12-1} =0.3012[/tex]

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

And replacing we got:

[tex]P(X \geq 2)= 1- [0.1422+0.3012]= 0.5566[/tex]

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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

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

[tex]X \sim Binom(n=12, p=0.15)[/tex]  

Part a

We want this probability:

[tex] P(X=0) = (12C0) (0.15)^0 (1-0.15)^{12-0} =0.1422[/tex]

Part b

We want this probability:

[tex] P(X=1) = (12C1) (0.15)^1 (1-0.15)^{12-1} =0.3012[/tex]

Part c

We want this probability:

[tex]P(X \geq 2)[/tex]

And we can use the complement rule:

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

And replacing we got:

[tex]P(X \geq 2)= 1- [0.1422+0.3012]= 0.5566[/tex]

Answer:

1364

Step-by-step explanation:

Answer:

It increases

Step-by-step explanation:

(5/x) + 5

As x decreases, 5/x increases.  So the expression increases.

Answer:

$12,884.08

Step-by-step explanation:

Assuming that interest is compounded annually, the future value of an invested amount 'P', at an interest rate 'r' for a period of 'n' years is given by the following equation:

[tex]FV = P*(1+r)^n[/tex]

Therefore, an investment of $8,000 at a rate of 10% per year for 5 years has a future value of:

[tex]FV = 8,000*(1+0.1)^5\\FV=\$12,884.08[/tex]

There will be $12,884.08 in the account after 5 years.

Answer:

Step-by-step explanation:

Assuming the interest was compounded annually, we would apply would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 8000

r = 10% = 10/100 = 0.1

n = 1 because it was compounded ince in a year.

t = 5 years

Therefore,

A = 8000(1+0.1/1)^1 × 5

A = 8000(1.1)^5

A = $12884.1

Answer:

Step-by-step explanation:

                    1

                1      1

            1      2      1

         1     3      3      1

      1    4     6       4       1

   1   5    10     10       5     1

 1    6   15   20      15    6     1

1    6   15   20      15    6     1

we use these for the expansion of (5x² + 2y³)⁶

1(5x²)⁶(2y³)⁰ + 6(5x²)⁵(2y³)¹ + 15(5x²)⁴(2y³)² + 20(5x²)³(2y³)³+  15(5x²)²(2y³)⁴+ 6(5x²)¹(2y³)⁵ + 1(5x²)⁰(2y³)⁶

78125ₓ¹²+187500ₓ¹⁰ y³ +37500ₓ⁸y⁶+20000ₓ⁶y⁹+6000x⁴y¹²+960x²y¹⁵+2y¹⁸

a.)a = 6, b = 9. the coefficient of xᵃyᵇ ( 20000ₓ⁶y⁹) = 20000

b) a = 2, b = 15. the coefficient of xᵃyᵇ ( 960x²y¹⁵) = 960

c) a = 3, b = 12. the coefficient of xᵃyᵇ is not present

d) a = 12, b = 0 the coefficient of xᵃyᵇ ( 78125ₓ¹²)  = 78125

e) a = 8, b = 9. the coefficient of xᵃyᵇ is not present

The coefficients of xᵃyᵇ for respective given values of a and b have been provided below.

We are given the expression;

(5x² + 2y³)⁶

Using online binomial expansion calculator gives us;

15625x¹² + 37500x¹⁰y³ + 37500x⁸y⁶ + 20000x⁶y⁹ + 6000x⁴y¹² + 960x²y¹⁵ + 64y¹⁸

We want to find the coefficient of xᵃyᵇ in the binomial expansion;

1) When a = 6 and b = 9, the coefficient is 20000

2) When a = 2, b = 15; the coefficient is 960

3) When a = 3 and b = 12; there is no coefficient

4) When a = 12 and b = 0; the coefficient is 15625

5) When a = 8 and b = 9; there is no coefficient.

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Answer: $553.20

Step by step explanation: Since they took the price down 29% multiply the price by 1.29 to add the 29 percent back to the total amount. 428.84 * 1.29 = 553.20

To find the probability of different scenarios involving defective components produced by a certain process, we can use the Poisson and binomial distributions.

(a) To find the probability that exactly 15 defective components are produced, we can use the formula for the Poisson distribution:

P(X=k) = (e^(-λ) * λ^k) / k!

Here, λ is the mean number of defective components produced in one day, which is 20. So, λ = 20. Substituting this value into the formula, we get:

P(X=15) = (e^(-20) * 20^15) / 15!

Calculating this expression will give us the probability.

(b) To find the probability that exactly 10 of the 15 defective components are repairable, we can use the binomial distribution since each defective component has a fixed probability of being repairable. Here, the number of trials is 15, and the probability of success (being repairable) is 0.60. Substituting these values into the binomial distribution formula, we can calculate the probability.

(c) Given the value of N, the number of defective components produced, X has a binomial distribution since X represents the number of repairable defective components. The probability of each component being repairable is constant, so it follows a binomial distribution.

(d) To find the probability that exactly 15 defective components are produced, with exactly 10 of them being repairable, we can multiply the probabilities obtained from parts (a) and (b) together, since these events are independent. Multiplying the results will give us the desired probability.

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(a) The probability that exactly 15 defective components are produced is 0.0516

(b) Given that exactly 15 defective components are produced, the probability that exactly 10 of them are repairable is 0.1241.

(c) Given the value of N, X follows a binomial distribution

(d) The probability that exactly 15 defective components are produced, with exactly 10 of them being repairable is 0.0064.

(a) Probability of Exactly 15 Defective Components

A Poisson distribution with mean λ = 20 is used. The formula is:

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

For k = 15 and λ = 20:

[tex]P(X = k) = (e^{(-20)} * 20^15) / 15! \approx 0.0516[/tex]

(b) Probability that Exactly 10 Out of 15 Defective Components are Repairable

This scenario uses a binomial distribution.

Given N = 15 defectives, the probability that exactly 10 are repairable (with p = 0.60) is:

[tex]P(X = 10 | N = 15) = C(15,10) * 0.6^{10} * 0.4^{5} \approx 0.1241[/tex]

(c) Distribution of X Given N

Given N = n

X (number of repairable components) follows a binomial distribution Bin(n, 0.60).

So, X | N = n follows Bin(n, 0.60).

(d) Probability of 15 Defective Components with Exactly 10 being Repairable

The joint probability is the product of the Poisson and binomial probabilities:

[tex]P(X = 15) * P(Y = 10 | X = 15)= 0.0516 * 0.1241 \approx 0.0064[/tex]

Answer:

0.0904 or 9.04%

Step-by-step explanation:

Mean glucose (μ) = 93.5

Standard deviation (σ) = 19.8

In a normal distribution, the z-score for any glucose value, X, is given by:

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

For X = 120, the z-score is:

[tex]Z= \frac{120-93.5}{19.8}\\ Z=1.3384[/tex]

A z-score of 1.3384 corresponds to the 90.96th percentile of a normal distribution. Therefore, the probability that glucose exceeds 120 in this population is:

[tex]P(X>120) = 1-0.9096=0.0904 = 9.04\%[/tex]

Answer:

Probability that glucose exceeds 120 in this population is 0.09012.

Step-by-step explanation:

We are given that in a population of adults between the ages of 18 and 49, glucose follows a normal distribution with a mean of 93.5 and standard deviation of 19.8, i.e.; [tex]\mu[/tex] = 93.5  and  [tex]\sigma[/tex] = 19.8 .

Let X = amount of glucose i.e. X ~ N([tex]\mu = 93.5 , \sigma^{2} = 19.8^{2}[/tex])

Now, the Z score probability is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

So,Probability that glucose exceeds 120 in this population =P(X>120)

P(X > 120) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{120-93.5}{19.8}[/tex] ) = P(Z > 1.34) = 1 - P(Z <= 1.34)

                                                   = 1 - 0.90988 = 0.09012 .

Answer: The slope of the tangent line at density 200 is -1.67kg/bean stem

Step-by-step explanation: The attached graph file shows the relationship between the yield of a certain crop f(×) as a function of the density of aphid x.

The second attached file shows the solution.

Answer:

(a) The Nyquist rate for the given signal is 2 × 5000π = 10000π. Therefore, in order to be able

to recover x(t) from xp(t), the sampling period must at most be Tmax =

2π

10000π = 2 × 10−4

sec. Since the sampling period used is T = 10−4 < Tmax, x(t) can be recovered from xp(t).

(b) The Nyquist rate for the given signal is 2 × 15000π = 30000π. Therefore, in order to be able

to recover x(t) from xp(t), the sampling period must at most be Tmax =

2π

30000π = 0.66×10−4

sec. Since the sampling period used is T = 10−4 > Tmax, x(t) cannot be recovered from

xp(t).

(c) Here, Im{X(jω)} is not specified. Therefore, the Nyquist rate for the signal x(t) is indeterminate. This implies that one cannot guarantee that x(t) would be recoverable from xp(t).

(d) Since x(t) is real, we may conclude that X(jω) = 0 for |ω| > 5000. Therefore, the answer to

this part is identical to that of part (a).

(e) Since x(t) is real, X(jω) = 0 for |ω| > 15000π. Therefore, the answer to this part is identical

to that of part (b).

(f) If X(jω) = 0 for |ω| > ω1, then X(jω) ∗ X(jω) = 0 for |ω| > 2ω1. Therefore, in this

part, X(jω) = 0 for |ω| > 7500π. The Nyquist rate for this signal is 2 × 7500π = 15000π.

Therefore, in order to be able to recover x(t) from xp(t), the sampling period must at most

be Tmax =

2π

15000π = 1.33 × 10−4

sec. Since the sampling period used is T = 10−4 < Tmax,

x(t) can be recovered from xp(t).

(g) If |X(jω)| = 0 for |ω| > 5000π, then X(jω) = 0 for |ω| > 5000π. Therefore, the answer to

this part is identical to the answer of part (a).

Answer:

the present value of the lease obligation for lease A will be $1,702,712.74 while the present value of lease B will be $1,764,741.73

Step-by-step explanation:

the present value of annual lease payment will be determined by discounting the total amount to be received from lease payment over years.

for lease A  , Pv =   A [tex]\fracA{1 -(1+r)^{-n} }{r}[/tex]

                         =   $200,000( 1 - (1+0.1)[tex]^{20}[/tex]) / 0.1 =  $1,702,712.74

for lease B , Pv =   $220,000( 1 - (1+0.1)[tex]^-{17}[/tex]) / 0.1  =  $1,764,741.73

the amount to be dicloased in the balance sheet at 31 december, 2018

The present value of lease payments assesses the current value of future lease liabilities, using a specified interest rate. Calculations for Lease A and Lease B would involve discounting their respective future lease payments at the given 10% interest rate to understand the company's financial obligations.

Present Value of Lease Payments

The present value of lease payments is a critical financial measure used by companies to assess the value of lease liabilities on their balance sheets. Generally, it accounts for all lease payments, discounted by a specific interest rate that reflects the time value of money. For Lease A, the company will make 20 annual payments of $200,000 each, starting on January 1, 2019. For Lease B, the company will make 17 annual payments of $220,000 each, starting on January 1, 2022. Using a discount rate of 10%, the present value of these lease payments can be calculated using the formula for the present value of an annuity.

For example, if a firm borrows $10,000 at an annual interest rate of 10%, it will owe $11,000 after one year because the original amount will accumulate interest. This illustrates the principle that a dollar today is worth more than a dollar in the future due to its potential to earn interest. Accordingly, the present value calculations for Lease A and Lease B will provide the amount the company should theoretically be willing to pay today to cover the lease payments over the course of the respective lease terms.

(a) Probability of upgrading in less than 12 minutes is practically 0. (b) Approximately 66 new files are required for 95% completion in less than 10 minutes.

Part (a):

First, let's calculate the mean and standard deviation of the time it takes to install all 68 files.

Given:

Mean time to install one file: [tex]$\mu = 15 , \text{sec}$[/tex]

Variance: [tex]$\sigma^2 = 11 , \text{sec}^2$[/tex]

The total time T to install all 68 files is the sum of 68 random variables, each with mean [tex]\mu$ and variance $\sigma^2$[/tex]. Since each installation time is independent, the mean of the total time is the sum of the individual means, and the variance of the total time is the sum of the individual variances.

Mean of total time:

[tex]\mu_{\text {total }}=68 \times \mu=68 \times 15 \mathrm{sec}=1020 \mathrm{sec}[/tex]

Variance of total time:

[tex]\sigma_{\text {total }}^2=68 \times \sigma^2=68 \times 11 \mathrm{sec}^2=748 \mathrm{sec}^2[/tex]

Standard deviation of total time:

[tex]\sigma_{\text {total }}=\sqrt{\sigma_{\text {total }}^2}=\sqrt{748} \mathrm{sec} \approx 27.34 \mathrm{sec}[/tex]

Now, to find the probability that the whole package is upgraded in less than 12 minutes (720 seconds), we convert this time into seconds and then use the cumulative distribution function of the normal distribution.

[tex]Z=\frac{X-\mu_{\text {total }}}{\sigma_{\text {total }}}[/tex]

Where:

X=720 (time in seconds)

[tex]\mu_{\text {total }}[/tex] =1020 (mean time in seconds)

[tex]\sigma_{\text {total }}[/tex] = [tex]\sqrt{748[/tex] (standard deviation in seconds)

Substituting the values:

[tex]\begin{aligned}& Z=\frac{720-1020}{27.34} \\& Z \approx-7.316\end{aligned}[/tex]

Now, we look up this Z value in a standard normal distribution table or use software to find the corresponding probability. The probability we seek is P(Z<−7.316).

Using a calculator or statistical software, this probability is extremely close to 0 (practically 0). Hence, the probability that the whole package is upgraded in less than 12 minutes is very close to 0.

Part (b):

Given that 95% of the time upgrading takes less than 10 minutes (600 seconds), we can use the same approach as above to find the corresponding Z value.

[tex]\begin{aligned}&Z=\frac{600-1020}{27.34}\\&Z \approx-14.468\end{aligned}[/tex]

To find the corresponding percentile from the standard normal distribution, we look up Z = -14.468 and find the corresponding percentile, which should be close to 0.05.

Now, let's find the number of files N required to ensure that 95% of the time the upgrading takes less than 10 minutes. We want the probability of completing the installation in less than 10 minutes to be 0.95, which corresponds to the Z value of approximately -1.645 (from standard normal distribution tables).

[tex]Z=\frac{X-\mu_{\text {total }}}{\sigma_{\text {total }}}=-1.645[/tex]

Solving for X:

[tex]\begin{aligned}& -1.645=\frac{X-1020}{27.34} \\& X=-1.645 \times 27.34+1020 \\& X \approx 978.52\end{aligned}[/tex]

So, the total time to install N files is approximately 978.52 seconds. Since we know the mean time to install one file is 15 seconds, we can find N by:

[tex]\begin{aligned}& N=\frac{978.52}{15} \\& N \approx 65.235\end{aligned}[/tex]

So, to ensure that 95% of the time upgrading takes less than 10 minutes, we would need to install approximately 65 new files. Since we can't install a fraction of a file, we would round up to the nearest whole number, giving us N = 66.

Complete Question:

Upgrading a certain software package requires installation of 68 new files. Files are installed consecutively. The installation time is random, but on the average, it takes 15 sec to install one file, with a variance of 11 square sec.

(a) What is the probability that the whole package is upgraded in less than 12 minutes?

(b) A new version of the package is released. It requires only N new files to be installed, and it is promised that 95% of the time upgrading takes less than 10 minutes. Given this information, compute N.

"You do not eat tofu" is derived using the Modus Tollens rule of inference.

The rule of inference that is used to derive the conclusion "You do not eat tofu" from the given statements is Modus Tollens.

Modus Tollens is a valid deductive argument form that follows this structure:

1. If P, then Q.

2. Not Q.

3. Therefore, not P.

The statements correspond to:

1. For all x, if x is healthy to eat, then x does not taste good.

2. Tofu is healthy to eat.

3. You only eat what tastes good.

Using these statements, we can infer:

1. If tofu is healthy to eat, then tofu does not taste good. (From statement 1)

2. Tofu does not taste good. (From statement 2 and the derived inference)

3. Therefore, you do not eat tofu. (Using Modus Tollens with statement 3 and the derived inference)

So, the conclusion "You do not eat tofu" is derived using the Modus Tollens rule of inference.

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Final answer:

The rule of inference used to derive the conclusion 'You do not eat tofu' is called Modus Tollens, which allows to deduce that if 'p implies q' is true and 'q' is false, then 'p' must also be false.

Explanation:

The student is asking about a logical inference rule used to derive a conclusion from a set of premises. Considering the provided statements, which can be summarized as:

For all x, if x is healthy to eat, then x does not taste good (All healthy foods taste bad).

Tofu is healthy to eat.

You only eat what tastes good.

We can see that the rule of inference used here is Modus Tollens. This rule of inference suggests that if we have a conditional statement (if p then q) and it is given that the consequent q is false (not q), then the antecedent p must also be false (not p).

To apply Modus Tollens:

Translate the given information into logical statements:
a) If something is healthy (p) then it does not taste good (q).
b) Tofu is healthy (p).
c) You do not eat what does not taste good (¬q).

Since tofu is healthy (p), it does not taste good as per the given rule (therefore, q is true). But the third statement says you do not eat what does not taste good, meaning (¬q). If (¬q) is valid, then by Modus Tollens, you do not eat tofu (¬p).

Thus, the usage of Modus Tollens allows us to infer that 'You do not eat tofu' from the given premises.

Answer:

  76.25°

Step-by-step explanation:

The solution to the differential equation is an exponential curve with a horizontal asymptote at Tm. It passes through (0, 145) and (30, 95), so the equation can be written as ...

  T = 80 +65((95-65)/(145-65))^(t/30)

  T = 80 +65(3/8)^(t/30)

That is, the temperature difference is reduced to 3/8 of its original value in 30 minutes.

Since the coffee in cup B cools twice as fast, it will cool to the same temperature (95°) in 15 minutes. In the next 15 minutes, the temperature difference will be reduced to (3/8)^2 of the original 80°, so will be 11.25°. That is, the temperature of cup B will be ...

  11.25° +65° = 76.25°

after 30 minutes.

The temperature of coffee in cup B after 30 minutes would be 45° F, as the rate of cooling is faster due to a larger exposed surface area.

According to the given condition, the rate of cooling (dT/dt) is proportional to the exposed surface area (S). It suggests that, with the surface area doubled for cup B, the rate of cooling (change in temperature) would also be double. Therefore, in the first 30 minutes, the temperature of coffee would decrease twice as fast as it did in cup A. In cup A, the temperature decreases from 145° F to 95° F in 30 minutes hence a decrease of 50° F. If the temperature of cup B decreases twice as fast, it decreases 100° F in 30 minutes. So, the temperature of the coffee in cup B after 30 minutes would be 145° F - 100° F = 45° F.

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Answer:

Independent Events

Step-by-step explanation:

Two events are mutually exclusive or disjoint if they cannot both occur at the same time.

Two events are independent if the probability of occurrence of one does not affect the probability of occurrence of the other.

The complement of any event B is the event [not B], i.e. the event that B does not occur.

Australia plays Argentina for the championship in the Rugby World Cup.

At the same time, Ukraine plays Russia for the World Team Chess Championship.

A = event that Argentina wins their rugby match.

B = the event that Ukraine wins their chess match.

The two events A and B are Independent as the outcome of one does not affect the outcome of the other