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maksim [4K]
3 years ago
6

Anton took out a 20-year loan for $90,000 at an APR of 10.5% compounded monthly, approximately what would be the total cost of h

is loan if he paid it off 12 years early?
Mathematics
2 answers:
snow_lady [41]3 years ago
5 0
The answer would be $159,661.69. i think
I am Lyosha [343]3 years ago
4 0

Answer: $159,661.69 ApeX

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Twelve shirts and three pairs of jeans cost $85.95 . Fifteen shirts and 2 pair of jeans costs $93.55.
Nesterboy [21]

Answer:

12s +3j= 85.95

15s +2j= 93.55

Step-by-step explanation:

First, define the variables:

Let the cost of a shirt be $s and the cost of a pair of jeans be $j.

From the first sentence:

12s +3j= 85.95 -----(1)

From the second sentence:

15s +2j= 93.55 -----(2)

5 0
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Which figure does not show symmetry?<br> A.<br> B.<br> C.<br> D<br> E.
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A. B. D. F.

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Evaluate each expression. (25^-3/2)^1/3
julsineya [31]

The result can be shown in multiple forms.

Exact Form:

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7 0
3 years ago
The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standar
34kurt

Answer:

a) 0.0081975

b) 0.97259

Step-by-step explanation:

The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

a. What is the probability that a line width is greater than 0.62 micrometer?

z = 0.62 - 0.5/0.05

z = 2.4

Probability value from Z-Table:

P(x<0.62) = 0.9918

P(x>0.62) = 1 - P(x<0.62)

= 0.0081975

b. What is the probability that a line width is between 0.4 and 0.63 micrometer?

For 0.4

z = 0.4 - 0.5/0.05

= -2

Probability value from Z-Table:

P(x = 0.4) = 0.02275

For 0.63

z = 0.63 - 0.5/0.05

= 2.6

Probability value from Z-Table:

P(x = 0.63) = 0.99534

P(x = 0.63) - P(x = 0.4)

= 0.99534 - 0.02275

= 0.97259

c. The line width of 90% of samples is below what value?

6 0
3 years ago
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