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Lana71 [14]
2 years ago
12

V=pir2h. Solve for pi

Mathematics
1 answer:
Fynjy0 [20]2 years ago
5 0

v = \pi {r}^{2} h \\ \\  \pi = \frac{v}{ {r}^{2}h }

I hope I helped you^_^

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Y=−2x+7<br> y=5x−7<br><br> x=<br> y=<br> ​
WINSTONCH [101]

Answer:

x = 2

y = 3

Explanation:

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2 years ago
A mixture of 17 %disinfectant solution is to be made from 10% and 20% disinfectant solution. How much of each solution should be
8090 [49]
Since 20 total gallons are needed, the amount of 10% solutions can be x and the amount of 20% solutions can be (20-x) since they add up to 20. In addition, the 17% solution has 20 gallons. Putting this into an equation, we get that 10% is 0.1 (by moving the decimal 2 spots to the left) , 0.2=20%, and 0.17=17%. With that information, we can determine that 10%*x+20%*(20-x)=20*0.17 due to that with 20 gallons of the 27% solution, we need a certain amount of solution. For every gallon, we add 0.17 of the solution, so we have 20*0.17. Similarly, we want that total amount to be represented by the amount of 20% + the amount of 10%.

0.1x+0.2(20-x)=0.1x+4-0.2x=-0.1x+4 using the distributive property
-0.1x+4=20*0.17=3.4. Subtract 4 from both sides to get -0.1x=-0.6, and multiply both sides by -10 (to separate x) to get x=6. Therefore, the amount of 10% solution is 6 gallons and the amount of 20% solution is 20-5=14 gallons
3 0
3 years ago
How do I write this in a word form number. 3.100? Thanks
mezya [45]
The answer would be:
    
       Three and one-tenths 
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4 0
3 years ago
Let X represent the full height of a certain species of tree. Assume that X has a normal distribution with a mean of 137.1 ft an
Rama09 [41]

Answer:

0.0668 = 6.68% probability that the height of a randomly selected tree is as tall as mine or shorter.

0.0228 = 2.28% probability that the full height of a randomly selected tree is at least as tall as hers.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 137.1, \sigma = 3.2

A tree of this type grows in my backyard, and it stands 132.3 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.

This is the pvalue of Z when X = 132.3. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{132.3 - 137.1}{3.2}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.0668 = 6.68% probability that the height of a randomly selected tree is as tall as mine or shorter.

My neighbor also has a tree of this type growing in her backyard, but hers stands 143.5 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.

This is 1 subtracted by the pvalue of Z when X = 143.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{143.5 - 137.1}{3.2}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the full height of a randomly selected tree is at least as tall as hers.

7 0
3 years ago
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