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Ede4ka [16]
3 years ago
5

Bad Investment After Mrs. Fisher lost 9% of her investment, she had $22,750. What was Mrs. Fisher's original investment?​

Mathematics
1 answer:
stira [4]3 years ago
7 0

Answer:

$2500

Step-by-step explanation:

Given that,

The lost percentage = 9%

Amount she had = $22,750

We need to find Mrs. Fisher's original investment.

As she had lost 9%, it means she will remain with 91% of her investment. Let the original investment is x.

So,

91\%\ of \ x=22,750\\\\\dfrac{91}{100}x=22,750\\\\x=\dfrac{22,750\times 100}{91}\\\\x=\$25000

So, her original investment is equal to $25000.

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the answer to your question is 807

Step-by-step explanation:

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Jamie's car is 9 yards long. Joey's car is 27 feet long. When you line them up one in front of the other, how long are they toge
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54

Step-by-step explanation:

9 x 3 = 27

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Using the equation representing the height of the firework (h = -16t2 + v0t + h0), algebraically determine the extreme value of
lubasha [3.4K]
-16t^2 + v0t + h0

=  -16(t^2 - 1/16 v0t)  + h0
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Read 2 more answers
Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much o
Art [367]

Answer:

The value will be used as the point estimate for the mean endowment of all private colleges in the United States is of 180.975 millions of dollars.

Step-by-step explanation:

When we have a sample, the mean of the sample is used as the point estimate for the mean of the entire population.

What value will be used as the point estimate for the mean endowment of all private colleges in the United States?

We have to find the mean of the 8 colleges. So

M = \frac{60.2 + 47.0 + 235.1 + 490.0 + 122.6 + 177.5 + 95.4 + 220.0}{8} = 180.975

The value will be used as the point estimate for the mean endowment of all private colleges in the United States is of 180.975 millions of dollars.

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3 years ago
The reading speed of second grade students in a large city is approximately normal, with a mean of 90 words per
kari74 [83]

Answer:

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

P(x⁻>96) =0.0359

Step-by-step explanation:

<em>Explanation</em>:-

<em>Given sample size 'n' =10</em>

<em>mean of the Population = 90 words per minute</em>

<em>standard deviation of the Population =10 wpm </em>

<em>we will use formula</em>

<em>                            </em>Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }<em></em>

<em>Let X⁻  = 96</em>

                           Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }

                          Z =  1.898

<em>The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute</em>

<em></em>P(X^{-}>x^{-} ) = P(Z > z^{-} )<em></em>

<em>                    = 1- P( Z ≤z⁻)</em>

<em>                    = 1- P(Z<1.898)</em>

                   = 1-(0.5 +A(1.898)

                   = 0.5 - A(1.898)

                   = 0.5 -0.4641 (From Normal table)

                  = 0.0359

<u><em>Final answer</em></u>:-

The probability that a random sample of 10 second grade students from  

                = 0.0359

4 0
3 years ago
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