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lawyer [7]
3 years ago
15

jacob just had lunch at a restaurant and leaves a 18% tip. Choose all the expressions that will give his total cost in terms of

his bill (b).
Mathematics
1 answer:
Leokris [45]3 years ago
8 0

Answer:

1.18b

Step-by-step explanation:

Jacob had lunch at a restaurant and his bill was b.

Now, Jacob leaves 18% of his bill amount as a tip.

So, the tip amount at the rate of 18% of the bill amount will be \frac{18 \times b}{100} =0.18 \times b

Therefore, the total cost of Jacob for the lunch will be b + 0.18b = 1.18b. (Answer)

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The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age gro
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Answer:

a) Null hypothesis:p\leq 0.12  

Alternative hypothesis:p > 0.12  

b) z_{\alpha}=1.64

c) z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362  

d) p_v =P(z>1.362)=0.0866  

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

\hat p=\frac{134}{1000}=0.134 estimated proportion of young drivers ages 18 – 24 that had an accident

p_o=0.12 is the value that we want to verify

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic

p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%:  Null hypothesis:[tex]p\leq 0.12  

Alternative hypothesis:p > 0.12  

The statistic would be given by:

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

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

z_{\alpha}=1.64

Part c

For this case the statistic would be given by:

z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362  

Part d

The p value can be calculated with the following probability:

p_v =P(z>1.362)=0.0866  

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

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Step-by-step explanation:

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