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maksim [4K]
3 years ago
11

I need help with problems 14-15.

Mathematics
1 answer:
oee [108]3 years ago
7 0

For 14, the answer is >. You inversed it. You are right about question 15.


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Tony’s club is selling oranges to raise money. For every box they sell, they get 1 1/8 dollars profit. They have sold 75 boxes a
dmitriy555 [2]

Tony’s club is selling oranges to raise money. For every box they sell, they get 1 1/8 dollars profit. They have sold 75 boxes already. How many more boxes must they sell to raise 180 dollars?

Answer: We are given:

The amount of money Tony's club get for every box they sell =1 \frac{1}{8} =1.125 dollar

They amount of money Tony's club has raised by selling 75 boxes is:

75 \times 1.125=84.375 dollars

The amount of money Tony's club is required to raise = 180 dollars

The remaining need to be raised is :

180-84.375=95.625 dollars

Therefore, the number of more boxes to be sold are:

\frac{95.625}{1.125}=85

Hence, 85 more boxes they must sell to raise 180 dollars

8 0
3 years ago
Use complete sentences to describe the similarities and differences between an expression and an equation.
nirvana33 [79]
Which class ques is this
3 0
3 years ago
(8.88×10^4) + (7.48×10^4)
podryga [215]
1.636x10^5 after adding or 163600 depending on the form wanted
4 0
3 years ago
On a standardized test with a normal distribution the mean score was 67.2. The standard deviation was 4.6. What percent of the d
Reika [66]

Answer:

P ( -1 < Z < 1 ) = 68%

Step-by-step explanation:

Given:-

- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :

                         u = 67.2

                         s.d = 4.6

- The random variable (X) that denotes standardized test scores following normal distribution:

                         X~ N ( 67.2 , 4.6^2 )

Find:-

What percent of the data fell between 62.6 and 71.8?

Solution:-

- We will first compute the Z-value for the given points 62.6 and 71.8:

                          P ( 62.6 < X < 71.8 )

                          P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )

                          P ( -1 < Z < 1 )

- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:

                          P ( -1 < Z < 1 ) = 68%

                          P ( -2 < Z < 2 ) = 95%

                          P ( -3 < Z < 3 ) = 99.7%

6 0
3 years ago
The mass of a textbook is about 1.25 kilograms. about how many pounds is this?
krok68 [10]
1 kilogram = 2.205 pounds
1.25 kilograms x 2.2 pounds = 2.75 pounds approximately
4 0
4 years ago
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