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Dovator [93]
3 years ago
12

If frequency of being late is .15 how often late in 100

Mathematics
1 answer:
koban [17]3 years ago
8 0
0.15 * 100 = 15

Answer: 15 times in 100
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What is the answer to 56x76
Simora [160]

Answer:  The answer is 4,256

Step-by-step explanation:

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3 years ago
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Write a linear equation passing through points (-7, -4) and (-5, -6)
STatiana [176]
The answer is y = -x + 11
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3 years ago
Hello can someone please help me on this :)
Zielflug [23.3K]

Answer:

Step-by-step explanation:

1.

Find value of P that makes this true

4p-5=`9

Solve for P by isolating it on one side:

4p-5+5=9+5

4p/4=14/4

P=3.5

2.

Translate the Algebraic Expression:

A waiter earns $128 for 6 hours of work including $86 of tips.

Subtract tips from total:

128-86=42

Divide Value between hours.

42/6=7

$7 per hour

3.

Find the Value of X that makes this true

-4x+26=-2

Solve for X by isolating it on one side:

-4x+26-26=-2-26

-4x/-4=-28/-4

X=7

4.

What is the first correct step in solving the equation:

33-2x=31

Subtract 33 from both sides

HOPE I HELPED!

BRAINLIEST WOULD BE APPRECIATED!

6 0
3 years ago
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A rain gutter is to be made of aluminum sheets that are 16 inches wide by turning up the edges 90. See the illustration.
goblinko [34]

Based on the width of the aluminum sheets and the angle of the edges, the depth that allows maximum cross-sectional area is 4 inches.

<h3>How much can the maximum cross-sectional area?</h3>

Assuming that the depth is x, the area would be:

= x (16 - 2x)

= 16x - 2x²

The maximum of the parabola is:

x = -b / 2a

= - 16 / (2 x -2)

= -16 / -4

= 4 inches

Find out more on using a parabola to solve for height at brainly.com/question/2274171

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7 0
2 years ago
Find the distance from N to T.
Stells [14]

Answer:

The distance from N to T is :  \sqrt{29}   or   (square root of 29)

Step-by-step explanation:

From the given the diagram

  • N(-2, -1)
  • T(3, 1)

\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

\mathrm{The\:distance\:between\:}\left(-2,\:-1\right)\mathrm{\:and\:}\left(3,\:1\right)\mathrm{\:is\:}

=\sqrt{\left(3-\left(-2\right)\right)^2+\left(1-\left(-1\right)\right)^2}

=\sqrt{\left(3+2\right)^2+\left(1+1\right)^2}

=\sqrt{5^2+\left(1+1\right)^2}

=\sqrt{5^2+2^2}

=\sqrt{25+4}

=\sqrt{29}

Therefore, the distance from N to T is :  \sqrt{29}   or   (square root of 29)

6 0
4 years ago
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