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3 years ago

The mean of a set of credit scores is and . Which credit score is within a z-score of 3.3?

Mathematics

1 answer:

sveta [45]3 years ago

4 0

Answer:

720

Step-by-step explanation:

The mean of a set of credit scores is u=690 and o=14.

mean = 690 and standard deviation SD = 14

We use z- score formula

z= 3.3 given

Plug in the values and find out x

Multiply 14 on both sides

46.2 = x- 690

Now add 690 on both sides

x= 736.2

Credit score is 720 that comes under 736.2

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Answer:

Step-by-step explanation:

From the figure attached,

Point B has been dilated to form point B'.

B(3, 1) → B'(6, 2)

→ B'[(2 × 3), (2 × 1)]

Since rule for the dilation of a point (x, y) by a factor of k is,

B(x, y) → B'(kx, ky)

By comparing the coordinates k = 2 is the scale factor by which the point B has been dilated about the origin.

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Answer.

Step-by-step explanation:

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Lorico [155]

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3 years ago

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Step-by-step explanation:

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3 years ago

At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500

kaheart [24]

Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

$P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}$

i.e. $PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}$

Substituting the values, we get,

i.e. $PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}$

i.e. $22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}$

i.e. $22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}$

i.e. $22000=\frac{P\times [1-0.4735]}{0.00625}$

i.e. $22000=\frac{P\times 0.5265}{0.00625}$

i.e. $P=\frac{22000\times 0.00625}{0.5265}$

i.e. $P=\frac{137.5}{0.5265}$

i.e. $P=261.16$

Thus, the total lifetime cost to pay of the loans compounded annually = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

7 0

3 years ago

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