Please help!!!!<br> Multiply. √3⋅2√2⋅5√8⋅√18

Alex needs to buy compressed air to fill empty cylindrical tanks for his tire store. Each tank is 12 feet deep and has a radius

agasfer [191]

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Formula
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Volume = πr²h

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Find Volume of one tank
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Volume of 1 tank = πr²h
Volume of 1 tank = 3.14 x 6² x 12
Volume of 1 tank = 678.24 ft³

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Find Volume of 5 tanks
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Volume of 1 tank = 678.24
Volume of 5 tanks = 678.24 x 5
Volume of 5 tanks = 3391.2 ft³

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Answer: 3391.2 ft³
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4 0

3 years ago

Write an equation of the line in slope-intercept form. (Pls answer fast)<br><br> Y=

nydimaria [60]

Answer:

$y = - \frac{1}{4} x + 3$

6 0

3 years ago

At a restaurant, servers keep 75% of the tips from each table, and support staff keeps 25%. A $20 tip is left at a table. How mu

EleoNora [17]

The server keeps $15

75% of 20 =

.75 x 20 = 15

7 0

3 years ago

Read 2 more answers

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in

erica [24]

Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0

3 years ago

Me Jacques wrote a bonus problem on the board. If Gina correctly answers, she gets extra time to read. The problem is to write a

balandron [24]

She could write 18 is a multiple of 9 but im nt sure if that answer is correct

6 0

3 years ago

Please help!!!! Multiply. √3⋅2√2⋅5√8⋅√18