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likoan [24]
3 years ago
7

What is the surface area of the triangular prism?

Mathematics
1 answer:
horrorfan [7]3 years ago
6 0
C)132
All you do is find the area of each flat surface and add them together.
(rectangle) A=w×h
(triangle) A=w×h/2
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Explain why it is necessary to rename 4 1/2 if you subtract 3/4 from it
ra1l [238]
It is neccesary because you cannot subtract them without having a common denominator so you have to rename 4 1/2
4 0
3 years ago
A cylinder has a height of 12 inches and a radius of 1.5 inches. What is its voulme? Round your answer to the nearest tenth. Use
USPshnik [31]

Find the area of the top, which is a circle.

Area of a circle = pi x r^2

Area = 3.14 x 1.5^2 = 7.065

Now for the volume multiply the area of the top by the height:

7.065 x 12 = 84.78 cubic inches

Round to the nearest tenth: 84.8 cubic inches

8 0
3 years ago
Read 2 more answers
Help meeèeeeeeeeeeeee​
mamaluj [8]
The answer is B your welcome
3 0
3 years ago
Please answer quick!
marysya [2.9K]

Answer:

The correct option is B

(-1)(1/2)(-1)(1)

Step-by-step explanation:

First thing to notice is that there is are two brackets with negative values, which means that the result must also be positive (or have two brackets with negative values).

Looking at the options, we can screen out options A and D, they have three brackets with negative values, and can't be chosen.

It is now between options B and C.

Option B, by inspection is simply 1/2

Option C is 12/4 = 3

The problem itself is 12/35

12/24 = 1/2

12/36 = 1/3

Since 12/35 is about 1/3, it is closer to 1/2 than 3, so 1/2 is the best option.

6 0
3 years ago
The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assu
m_a_m_a [10]

Answer: 0.0170

Step-by-step explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e. \mu=23.50

\sigma=5

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let \overline{x} be the sample mean.

Formula : z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

Then, the probability that the sample mean is at least $25.00 will be :-

P(\overline{x}\geq\25.00)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\geq\dfrac{25-23.50}{\dfrac{5}{\sqrt{50}}})\\\\=P(z\geq2.12)\\\\=1-P(z

Hence, the likelihood the sample mean is at least $25.00= 0.0170

5 0
3 years ago
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