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marta [7]
3 years ago
14

During each minute of a comedy show, Carlin laughs 5 times. Carlin watches 3 comedy shows every day, and each show is 40 minutes

long. How many times does Carlin laugh every day due to the comedy shows?
Mathematics
2 answers:
Snowcat [4.5K]3 years ago
7 0

Answer: Carlin laughs 600 times every day due to the comedy shows.

Step-by-step explanation:

Given : During each minute of a comedy show, Carlin laughs 5 times.

Total shows she watches = 3

Length of each show = 40 minutes

Total length of the shows =(Total shows she watches) x (Length of each show)

=  3 x 40 units = 120 units

Number of times Carlin  laugh every day due to the comedy shows = (Total length of the shows) x  (Number of times she laughs in one minute )

= 120 x 5

= 600

Hence, Carlin laughs 600 times every day due to the comedy shows.

Salsk061 [2.6K]3 years ago
3 0
5 x 40 = 200
200 x 3 = 600

Carin laughs 600 times every day due to the comedy shows.
(That is a hell lot of laughing in one day... )
;)
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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.  

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Answer:

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Step-by-step explanation:

A clothes shop has some special offers. ?

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The cost of 2 T-shirts since it is a buy 1 get 1 free => £7.40

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The total cost = £7.40 + £9.80

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