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Lorico [155]
3 years ago
7

Alvin has a prism-like water tank whose base area is 0.90.90, point, 9 square meters and height is 0.60.60, point, 6 meters. He

wants to buy guppy fish, and the pet store owner tells him to make sure their density in the tank isn't more than 555 fish per cubic meter.
Mathematics
1 answer:
Helen [10]3 years ago
8 0

Answer:

2 fishes

Step-by-step explanation:

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Can someone help me with this? i'll award brainiest :D
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About 79 boys i think

Step-by-step explanation:

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The diagram shows three squares, PQRS, TUVW and WXYZ.
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44

Step-by-step explanation:62+74     -  180

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Solve the following equation for the
Akimi4 [234]
Divide -3
-3/-3a = 15/-3
a = -5
5 0
3 years ago
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Evaluate the factorial.<br> 11!
Dafna11 [192]

Answer:

The factorial of 11! is exactly 39916800

The number of trailing 0s in 11! is 2

The number of digits in 11 factorial is 8.

The factorial of 11 is calculated as below:

11! = 11 • 10 • 9 • 8 • 7 ... 3 • 2 • 1

Step-by-step explanation:

6 0
3 years ago
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To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil
Debora [2.8K]

Answer:

statistic value t= 5.43

p-value: < 0.00001.

Step-by-step explanation:

Hello!

To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.

The study variable is:

X: Max. penetration of a steel conduit.

The data of the sample

n= 45

sample mean X[bar]= 53.4

sample standard deviation S= 4.2

The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50

The hypothesis is:

H₀: μ ≤ 50

H₁: μ > 50

α: 0.05

To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.

Under the null hypothesis, the critical region and the p-value are one-tailed.

Critical value:

t_{n-1; 1-\alpha } = t_{44; 0.95} = 1.68

Rejection rule:

Reject the null hypothesis when t ≥ 1.68

t= \frac{53.4 - 50}{\frac{4.2}{\sqrt{45} } }

t= 5.43

The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.

p-value:

P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.

The p-value is less than α so the decision is to reject the null hypothesis.

Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.

I hope it helps!

4 0
3 years ago
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