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denis-greek [22]
2 years ago
9

A customer wants to tip 15% on a restaurant bill that is $35.99. How much gratuity did the customer leave?​

Mathematics
1 answer:
Anton [14]2 years ago
8 0
35.99 times .15 = about 5.40
The customer left $5.40 gratuity.
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What is the product of 73 and 4.4 x 10^6 expressed in scientific notation?​
Dima020 [189]

Answer: 7.3 x 10^1 & 4400000.

Step-by-step explanation: Count the place holders and just add zeros to the exponential amount.

7 0
3 years ago
4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) Statistical decision  

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 right tailed test the p value would be:  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

8 0
3 years ago
jaimie ran 3 1/2 on monday.She ran half as far on teusday as she did on monday.How far did jaimie run in all on monday and teusd
Soloha48 [4]

Answer: 5 1/4

Step-by-step explanation:

hope this helps!

4 0
2 years ago
URGENT 50 PTS
Len [333]
Point c

explanation:
6 0
2 years ago
Can you do this one pls
nordsb [41]

Answer:

its c

Step-by-step explanation:

6 0
3 years ago
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