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SashulF [63]
3 years ago
5

What is the unit rate? $101.25 earned in 9 hours

Mathematics
2 answers:
vekshin13 years ago
7 0

Answer:

$11.25

Step-by-step explanation:

Bumek [7]3 years ago
3 0

Answer:

$11.25

Step-by-step explanation:

Divide $101.25 by 9

11.25

So $11.25 were earned each hour.

Hope this helps :)

Please consider Brainliest :)

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Will mark brainliest
VMariaS [17]

Answer:

Interquartile range => 7

Third quartile => 22

Range => 13

First quartile => 15

Step-by-step explanation:

Order your data set, from the least amount to the highest amount:

$12, $14, $15, $15, $15, $15, $16, $22, $24, $25

Interquartile range (IQR) = third quartile (Q3) - first quartile (Q1)

Q1 = the middle value of the lower part of the data set, from the median to your left.

Q3 = the middle value of the upper part of the data set, from the median to your right.

The median lies between the 5th and 6th value that is enclosed in the parenthesis below:

$12, $14, ($15), $15, $15,[Median], $15, $16, ($22), $24, $25

The median divides the data set into upper and lower part.

Median = \frac{15 + 15}{2} = 15

First quartile: Q1 = $15

Third quartile: Q3 = $22

IQR = $22 - $15 = $7

Range = highest amount - least amount = 25 - 12 = $13

3 0
3 years ago
1.) Vivian combined different amounts of white paint, blue paint, and green paint to make 145 milliliters (ml) of paint of a new
Ivahew [28]
So here are the answers to the given questions above:
1. Since the equation is already given, what we are going to do is to solve for x first. 
<span>x + (2x – 10) + 1/2 (2x – 10) = 145
x + (2x -10) + x - 5 = 145
x + 2x + x - 10 - 5 = 145
4x = 145 +10 + 5
4x = 160 <<divide both sides by 4 and we get
x = 40.
Therefore, the amount of white paint is 40ml.
The amount of blue paint is 70ml and the green paint is 35ml.
So the difference between the a</span><span>mounts of white paint and blue paint Vivian combined is 30ml.
2. T</span>he equation of the line that passes through the points (-2, 1) and (1, 10) would be: <span>3x - y = -7, the last option. In order to check, just plug in the given ordered pairs.</span>
3 0
3 years ago
Help me !!!!!!!!!!!!!!!!!!!!
True [87]

Answer:

0 isnt a number loll

Step-by-step explanation:

B

6 0
3 years ago
Somebody please help me. This makes zero sense.
earnstyle [38]

Answer:

d

Step-by-step explanation:

rearrange it like an equation.

take everything else out of the middle apart from x.

make sure that what you do to the middle you do to both sides

6 0
2 years ago
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
3 years ago
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