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3 years ago

PLEASE HELP Find the surface area of the triangular prism.

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

139

Step-by-step explanation:

First, find all the faces that you must find the areas of. In this case, you need to find the areas of two triangles and three rectangles.

In this triangular prism, the base is one triangle. If the area of the base is 7.75, we can assume that that holds true for both bases, and multiply 7.75 by 2 in order to find the area of both triangles.

Now find the area of each of the three rectangles by multiplying their individual heights and bases by each other. You should get 38, 47.5, and 38 as the areas of the rectangles.

Now add all the individual area's together. (They're bolded for clarity).

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3 years ago

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Kyle's phone bill is $40 per month. How much does he have to pay for half a year of phone service? *

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Answer:

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Step-by-step explanation:

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3 years ago

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Tank a has a capacity of 9.5 gallons. 6 1/3 gallons of the tank's water are poured out. how many gallons of water are left in th

Advocard [28]

Answer:

Water left in tank is $3\frac{1}{6}$ gallons.

Step-by-step explanation:

Given :

Capacity of tank = 9.5 gallons $=\frac{95}{10}$

Water taken out of tank = $6\frac{1}{3}=\frac{19}{3}$ gallons

We have to find the water left in the tank.

Water left in the tank = Total capacity of tank - water taken out of tank

= $\frac{95}{10}-\frac{19}{3}$

Taking LCM of (3,10)= 30

$\Rightarrow \frac{95 \times 3-19 \times10}{30}$

Solving further, we get,

$\Rightarrow \frac{285-190}{30}$

$\Rightarrow \frac{95}{30}=\frac{19}{6}=3\frac{1}{6}$

Thus, Water left in tank is $3\frac{1}{6}$ gallons..

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3 years ago

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In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo

larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine $z_{\frac{\alpha}{2}}=z_{0.005}$

Now by using z score table we find that $z_{\frac{\alpha}{2}}=2.58$

The boundaries of the confidence interval are:

$\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64$