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Vitek1552 [10]
2 years ago
7

Solve for variable its in pre algebra

Mathematics
1 answer:
Natalka [10]2 years ago
7 0

Answer:

7

Step-by-step explanation:

7

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At a store ,apples cost $5.00 for 2 pounds.what is the ratio of dollars to pounds of apples
Firdavs [7]

Answer:

$2.50 per pound of apple

Step-by-step explanation:

Or if you have to write it as a fraction $2.50/ 1

8 0
2 years ago
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Which number is not a common factor of 32 and 48
Colt1911 [192]
5, 3, 7, 9 are all non-common factors
8 0
3 years ago
The first three towers in a sequence are shown. The nth tower is formed by stacking n blocks on top of an n times n square of bl
erik [133]

Answer:

The 99th tower contains 9900 blocks.

Step-by-step explanation:

From the question given, we were told that the nth tower is formed by stacking n blocks on top of an n times n square of blocks. This implies that the number of blocks in n tower will be:

n + n²

Now let us use the diagram to validate the idea.

Tower 1:

n = 1

Number of blocks = 1 + 1² = 2

Tower 2:

Number of blocks = 2 + 2² = 6

Tower 3:

Number of blocks = 3 + 3² = 12

Using same idea, we can obtain the number of blocks in the 99th tower as follow:

Tower 99:

n = 99

Number of blocks = 99 + 99² = 9900

Therefore, the 99th tower contains 9900 blocks.

7 0
3 years ago
Write two equivalent expressions that represent the rectangular array below.
Ainat [17]
Area formula for rectangles: L x W

Answer: 3 x 2a + 3 x 5 = 3 x 2a x 5

You know that the expressions are equivalent, because when you solve for a you get a = 5/8 or 0.625
3 0
2 years ago
NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud
SSSSS [86.1K]

Answer:

The large sample n = 190.44≅190

The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given  population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }

The 85% confidence level z_{\alpha } = 1.44

\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}

\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

7 0
3 years ago
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