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jonny [76]
3 years ago
14

Gus has a fish tank that holds 471047104710 inches^3 3 start superscript, 3, end superscript of water. He is using a cylinder sh

aped bucket with a radius of 555 inches and a height of 202020 inches to fill the tank. How many times will Gus need to fill the bucket to completely fill the fish tank if he doesn't spill a drop? Use 3.143.143, point, 14 for \piπpi.
Mathematics
1 answer:
lilavasa [31]3 years ago
7 0
The first thing we must do for this case is find the volume of the bucket.
 We have then that by definition the volume of a cylinder is given by:
 V = pi * r ^ 2 * h
 Where,
 r: radio
 h: height
 Substituting the values we have:
 V = (3.14) * (5) ^ 2 * (20)
 V = 1570 in ^ 3
 Then, the number of times the bucket should be filled will be given by:
 N = (4710) / (1570)
 N = 3 times
 Answer:
 
Gus will need to fill the bucket to completely fill the fish tank if it does not spill a drop about:
 
N = 3 times
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Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Proportion of 0.6

This means that p = 0.6

Sample of 46

This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.6}{0.0722}

Z = -1.38

Z = -1.38 has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

8 0
3 years ago
I NEED HELP PLEASE ! :)
dedylja [7]

We can write a proportion to resemble the problem;

AE/ED = AB/BC

AE = 9

ED = 6

AB = x

BC = 10

Substitute with the given values.

9/6 = x/10

9/6 * 10 = x/10 * 10

90/6 = x

15 = x

Therefore, the answer is 15.

Best of Luck!

7 0
3 years ago
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The sum of two times a number and -2 is at least 8<br><br> how can I solve it plz help
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To solve, your equation would be 2x - 2 >/= 8. You add 2 to both sides, getting 2x >/= 10. Then divide by two on both sides, your answer would be x >/= 5. (x is greater than or equal to 5).
4 0
3 years ago
Two friends sold many pieces of furniture and made ​$1550 during their garage sale. They had fifteen more​ $10 bills than​ $50 b
Novosadov [1.4K]

Answer:

The number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

Step-by-step explanation:

Let n₁ = number of $ 10 bills,

Let n₂ = number of $ 20 bills and

Let n₃ = number of $ 50 bills

Given that 10n₁ + 20n₂ + 50n₃ = 1550    (1)

Also, since we have fifteen more​ $10 bills than​ $50 bills, n₁ = n₃ + 15  (2)

and we have 4 more than three times as many​ $20 bills as​ $50 bills. n₃ = 3n₂ + 4. (3)

substituting equations (2) and (3) into (1), we have

10(n₃ + 15) + 20n₂ + 50n₃ = 1550

expanding the bracket, we have

10n₃ + 150 + 20n₂ + 50n₃ = 1550

collecting like terms, we have

60n₃ + 150 + 20n₂ = 1550

inserting equation (3), we have

60(3n₂ + 4) + 150 + 20n₂ = 1550

expanding the bracket, we have

180n₂ + 240 + 150 + 20n₂ = 1550

collecting like terms, we have

200n₂ + 390 = 1550

subtracting 390 from both sides, we have

200n₂ = 1550 - 390

200n₂ = 1160

dividing both sides by 200, we have

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n₂ ≅ 6 since it cannot be a fraction.

Substituting this into (3), we have

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substituting n₃ into (2), we have

n₁ = n₃ + 15 = 22 + 15 = 37

So, the number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

3 0
3 years ago
60+6=48<br> What is the value of b?
mina [271]

Answer:

-2

Step-by-step explanation:

Assuming you mean 60 + 6b = 48

subtract 60 from both sides: 6b = 48 - 60 = -12

Now divide both sides by 6 to solve for b: b = -12/6 = -2

3 0
3 years ago
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