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monitta
3 years ago
9

Can someone please help me solve the problem for my math homework

Mathematics
2 answers:
stepan [7]3 years ago
8 0
17.5 cubic inches... it's literally height multiplied by the area of the base...
Shtirlitz [24]3 years ago
6 0
The answer is 15.25 cubic inches
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I need help with that one please
k0ka [10]
Your answers is correct Don’t worry good job
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3 years ago
I would like for some to help me because I really need help
torisob [31]

Answer:

whats the question

Step-by-step explanation:

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3 years ago
In the past, the mean running time for a certain type of flashlight battery has been 9.7 hours. The manufacturer has introduced
Nadya [2.5K]

Answer:

Step-by-step explanation:

From the information given:

\mathbf{H_o: \mu = 9.7 \ hours}

\mathbf{H_1: \mu > 9.7 \ hours}

The type 1 error is rejecting \mathbf{H_o} when

The meaning of Type 1 error is rejecting the claim that the mean running time is 9.7 hours when actually the mean running time is greater than 9. 7 hour.

6 0
3 years ago
If the width of a pool is 60 feet and the perimeter is 200 feet, how long is the pool?
velikii [3]

Answer:

140

Step-by-step explanation:

200-60 = 140

8 0
3 years ago
Read 2 more answers
A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
alukav5142 [94]

Answer:

0.17 °/s

Step-by-step explanation:

Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

We differentiate the above expression with respect to time to have

dsinθ/dt = d(D/L)/dt

cosθdθ/dt = (1/L)dD/dt

dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]

dθ/dt = (1/Lcosθ)dD/dt

dθ/dt = (1/L√[1 - (D/L)²])dD/dt

dθ/dt = (1/√[L² - D²])dD/dt

Substituting the values of the variables, we have

dθ/dt = (1/√[20² - 16²]) 2 ft/s

dθ/dt = (1/√[400 - 256]) 2 ft/s

dθ/dt = (1/√144) 2 ft/s

dθ/dt = (1/12) 2 ft/s

dθ/dt = 1/6 °/s

dθ/dt = 0.17 °/s

8 0
3 years ago
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