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hichkok12 [17]
2 years ago
9

Mason coaches a swim team for $15 per hour and works at a printing shop for $18 per hour. He wants to earn at least $125per week

but cannot work for more than 20 hours a week. Writes a graph linear system of inequalities to represent this situation. Determine one possible solution to Mason's problem. Let the number of hours coaching = x and the number of hours at the printing shop = y.
Mathematics
1 answer:
skelet666 [1.2K]2 years ago
7 0

Answer:

13

Step-by-step explanation:

step-by-step explanation: 169-1=168168-3=165165-5=160160-7=153153-

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What is the product (3 x 10') * (2 x 10") in scientific
Mariana [72]

Answer:

600

Step-by-step explanation:

3 x 10 = 30

2 x 10 = 20

30 x 20 = 600

5 0
3 years ago
40 points fast For the following set of test scores, identify:
myrzilka [38]

Answer:

 The measure of center would this student want his teacher to use is the

median= {\bf 75}

Mean= {\bf 5}  ,  Mode={\bf 82}  , Range={\bf40}  and   Interquartile Range= {\bf 32}

Step-by-step explanation:

Given Data set= 50,55,70,75,82,82,90

Number of elements in data set= 7

To find Mean

The ‘Mean” is the average of a set of numbers.

The "Mean" is computed by adding all of the numbers in the data together and dividing by the number  of elements contained in the data set.

Mean= \frac{50 +55 +70 + 75 + 82 + 82 + 90} {7 }

Mean= \frac{504}{7}

Mean= 5

Median

The “Median” is the middle value of a set of ordered numbers.

Therefore Median=75

Mode

The "Mode" for a set of data is the value that occurs most often.

It is not uncommon for a data set to have more than one mode. This happens when two or more  elements occur with equal frequency in the data set.

Therefore Mode=82

Range

The "Range" is the difference between the largest value and smallest value in a set of data.

Range= 90-50

Range= 40

Interquartile Range

The “Interquartile Range” is the difference between smallest value and the largest value of the middle  50% of a set of data.

The "Interquartile Range" is from Q1 to Q3:

To find the interquartile range of a set of data:

The cut the list into four equal parts

. The quartiles are the “cuts”

The interquartile range is the distance between the two middle sets of data

Interquartile Range= Q_3-Q_1

Interquartile Range= 82-50

Interquartile Range= 32

5 0
3 years ago
HELP GIVEING BRAINLIST What is the solution set of -11 > 1 + a?
Alja [10]

Answer:

the solution set is -11 > 1 + a?

a > -10

a -12

a < -10

3 0
3 years ago
Read 2 more answers
A circle has a radius of 8 inches. Find the area of a sector of the circle if the sector has an arc that measures 45°.
alexgriva [62]
To look for the area of a sector of a circle, the following formula is used:

Area = ( n / 360 ) * pi * r^2

Where: n = measure of arc in degrees
pi = 3.1416
r = radius

Since we are already given all of the needed parts of the formula, direct substitution is done, as shown below:

Area = ( n / 360 <span>) * pi * r^2
</span>Area = ( 45 / 360 ) *3.1416 * (8)^2
Area = 25.133 in^2

Therefore, the area of the sector of the circle is 25.133 sq. in.
5 0
3 years ago
Read 2 more answers
An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the
iragen [17]

Answer:

A. P("Select 2 black balls and then select 2 white balls")=35/768

Step-by-step explanation:

A. First, we have to get an idea of what our experiment talks about:

If I take a black ball, then we note its color and put it back to the urn with 2 more balls of the same color.

So, every time that we take a ball the quantity of balls in the urn changes with the probability that we had before replacing. The first case (that the first ball you select is black) would go like this:

P("Select a black ball first try") = 7 (number of black balls)/12 (total of balls in the urn)

P(B₁)=7/12

Then, the amount of balls in the urn rise to 14 (12 + 2 balls added) and now the quantity of black balls it´s 9 (7 + 2 added). The urn currently has 9 black balls and 5 white balls, of course:

P("Select black ball second try after getting a black ball before") = 9 (number of black balls)/ 14 (total of balls in the urn)

P(B₁B₂)=9/14

<u>Notice that while we keep taking black balls, the quantity of white ball keeps constant</u> and, for the probability that we take a white ball, we just consider the change in the number of balls in the urn

P(B₁B₂W₃)=5/16

And now, we consider the new white balls that are put in the urn (2 white balls, for a total of 18 and 7 white balls):

P(B₁B₂W₃W₄)=7/18

Because all these probabilities are chained and everything comes from the first ball we choose <u>(if you use a tree diagram, they will be in the same branch</u>), we multiply the probabilities to find the probability that all happens in the same experiment:

P(B₁)*P(B₁B₂)*P(B₁B₂W₃)*P(B₁B₂W₃W₄) = 7/12 * 9/14 * 5/16 * 7/18 = 2205/48384

P(B₁)*P(B₁B₂)*P(B₁B₂W₃)*P(B₁B₂W₃W₄)=35/768

And this is the final answer

7 0
3 years ago
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