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yawa3891 [41]
3 years ago
14

Marcus hired a car service to drive him to the airport. The bill came to $56.80. He gave the driver a 20% tip. How much money di

d he give the driver as a tip?
Mathematics
2 answers:
GenaCL600 [577]3 years ago
6 0
Since the tip is 20% of the bill so it is $11.36. So he gave $11.36 as tip
aleksandr82 [10.1K]3 years ago
4 0
He gave the driver a tip of $11.36

58.80*0.20 = 11.36
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Find the 5th term of the sequence defined by the give rule. f(n) = n²+ 5. A step by step explanation with answer would be greatl
skad [1K]

Answer:

<h3>The 5th term is 30</h3>

Step-by-step explanation:

Given the formula

f(n) = n² + 5

where n is the number of terms

So from the question we were told to find the 5th term that's

n = 5

In order to find the 5th term substitute the value of n that's 5 into the rule

We have

f(5) = 5² + 5

= 25 + 5

= 30

<h3>f(5) = 30</h3>

So the 5th term of the sequence in the given rule is 30

Hope this helps you

3 0
4 years ago
Can someone help me in this please
Karolina [17]
Is the value of b is 5?
5 0
3 years ago
Guest ages at a ski mountain resort typically have a right-skewed distribution. Assume the standard deviation (σ) of age is 14.5
ValentinkaMS [17]

Answer: (30.49 years, 42.31 years)

Step-by-step explanation:

The formula to find the confidence interval is given by :-

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}.

, where \overline{x} = Sample mean

z* = Critical value.

\sigma = Population standard deviation.

n= Sample size.

As per given , we have

\overline{x}=36.4

\sigma=14.5

n= 40

We know that the critical value for 99% confidence interval : z* = 2.576 (By z-table)

A 99 percent confidence interval for µ, the true mean age of guests will be :

36.4\pm (2.576)\dfrac{14.5}{\sqrt{40}}\\\\ 36.4\pm (2.576)2.29265130362\\\\=36.4\pm5.90586975813\\\\\approx36.4\pm5.91\\\\=(36.4-5.91,\ 36.4+5.91)\\\\=(30.49,\ 42.31)

∴ a 99 percent confidence interval for µ, the true mean age of guests  = (30.49 years, 42.31 years)

7 0
3 years ago
What would be the critical value for a two-tailed, one-sample t test with 26 degrees of freedom (df = 26) and an alpha level (p
Alika [10]

Answer: The critical value for a two-tailed t-test = 2.056

The critical value for a one-tailed t-test = 1.706

Step-by-step explanation:

Given : Degree of freedom : df= 26

Significance level : \alpha=0.05

Using student's t distribution table , the critical value for a two-tailed t-test will be :-

t_{\alpha/2, df}=t_{0.025,26}=2.056

The critical value for a two-tailed t-test = 2.056

Again, Using student's t distribution table , the critical value for a one-tailed t-test will be :-

t_{\alpha, df}=t_{0.05,26}=1.706

The critical value for a one-tailed t-test = 1.706

6 0
3 years ago
What is the equation of the graph below?
alex41 [277]

Answer:

A

Step-by-step explanation:

8 0
2 years ago
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