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FinnZ [79.3K]
3 years ago
9

A sneaker store salesman had $4,125 in total monthly sales last month. He made $165 in commission from those sales. What is the

salesman's commission as a percent of his total monthly sales? Enter your answer in the box.
Mathematics
1 answer:
AysviL [449]3 years ago
8 0
His total commission is 4%
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80 degrees over the course of four months
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What is the slope intercept for that has the point (2,2) and is parallel to y=x+4?
gulaghasi [49]
The new line will have the same slope as the old one.
y = x + 4
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3 years ago
PLSSSSSSSSSSSSSS HELP <3
Komok [63]

1. Given

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3 years ago
You buy books for 64 cents each, you plan to sell them for 75 cents each. You're left with 100 unsold books, and you lost $12. H
PSYCHO15rus [73]

The number of books sold is 473.

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

  • The original cost of each book = $0.64
  • The selling price of each book = $0.75

The difference between the original price and selling price of the book gives the profit per book.

The profit of one book = Selling price - Original price

Let,

  • The total number of books be 'x'.
  • The number of books sold be 'y'.
  • The unsold books is 100.
  • The total profit is -12 because it was gone to a loss of $12.

Therefore, the equation is formed as

total Profit = 0.75y - 0.64x

⇒ 0.75y - 0.64x = -12 --------(1)

Total books = sold books + unsold books

x = y + 100

⇒ x-y = 100  -------(2)

Substitute x= 100+y in the eq(1),

0.75y - 0.64(100+y) = -12

0.75y - 64 -0.64y = -12

0.11y = -12 +64

y = 52 / 0.11

y = 472.7

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The number of book sold is 473 books.

The total number of books is (100+473) = 573 books.

8 0
3 years ago
2. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of
IgorC [24]

Answer:

a) 37.31 b) 42.70 c) 0.57 d) 0.09

Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., \mu = 35 and \sigma = 6. We know that the probability density function for a normal distribution with a mean of \mu and a standard deviation of \sigma is given by

f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]

in this case we have

f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]

Let X be the random variable that represents a row score, we find the values we are seeking in the following way

a)  we are looking for a number x_{0} such that

P(X\leq x_{0}) = \int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65, this number is x_{0}=37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number  x_{1} such that

P(X\leq x_{1}) = \int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9, this number is x_{1}=42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

P(28\leq X\leq 38)=\int\limits^{38}_{28} {f(x)} \, dx = 0.57

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

P(41\leq X\leq 44)=\int\limits^{44}_{41} {f(x)} \, dx = 0.09

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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