What is the quotient of n and 8

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal. During one

Triss [41]

Answer:

0.82 = 82% probability that at least one party will not order drinks

Step-by-step explanation:

For each party, there are only two possible outcomes. Either they will order drinks with their meal, or they will not. The probability of a party ordering drinks with their meal is independent of other parties. So the binomial probability distribution is used 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal.

This means that $p = 0.75$

During one hour, Ryan served 6 parties. Assuming that each party is equally likely to order drinks, what is the probability that at least one party will not order drinks?

6 parties, so n = 6.

Either all parties will order drinks, or at least one will not. The sum of the probabilities of these events is decimal 1. So

$P(X = 6) + P(X < 6) = 1$

We want P(X < 6). So

$P(X < 6) = 1 - P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.18$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.18 = 0.82$

0.82 = 82% probability that at least one party will not order drinks

5 0

3 years ago

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Answer the question

Artist 52 [7]

whats your question?

5 0

3 years ago

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The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by K(x) 5x/x^2

Olegator [25]

Given :

The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by $K(x) = \dfrac{5x}{x^2+9}$ .

To Find :

Find the time at which the concentration is a maximum. b. Find the maximum concentration.

Solution :

For maximum value of x, K'(x) = 0.

$K'(x) = \dfrac{5(x^2+9)- 5x(2x)}{(x^2+9)^2}=0\\\\5x^2+45-10x^2=0\\\\5x^2 = 45\\\\x = \pm 3$

Since, time cannot be negative, so ignoring x = -3 .

Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6

Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.

Hence, this is the required solution.

5 0

2 years ago

Which data set could be represented by the box plot shown below?

Lapatulllka [165]

Answer:

The answer is C

Step-by-step explanation:

It starts at 25 and then it stops at 38.

I hope this helps!

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2 years ago

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In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders o

melomori [17]

Answer:

A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B. Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

C. $z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D. $z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

E. Fail to the reject the null hypothesis

F. So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ 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 true proportion of inaccurate orders is not significantly different from 0.1.

Step-by-step explanation:

Data given and notation

n=362 represent the random sample taken

X=33 represent the number of orders not accurate

$\hat p=\frac{33}{363}=0.0912$ estimated proportion of orders not accurate

$p_o=0.10$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

A: Write the claim as a mathematical statement involving the population proportion p

We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B: State the null (H0) and alternative (H1) hypotheses

Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

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

C: Find the test statistic

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

$z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D: Find the critical value(s)

Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

$z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

P value

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

E: Would you Reject or Fail to Reject the null (H0) hypothesis.

Fail to the reject the null hypothesis

F: Write the conclusion of the test.

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ 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 true proportion of inaccurate orders is not significantly different from 0.1.

6 0

3 years ago