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Maslowich
3 years ago
14

The length of a rectangle is 3 meters more than the width. If the perimeter is 170 meters, what are the length and the width?

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

41,44

Step-by-step explanation:

make the width x. lenght = x+3.

so, the perimeter will be x+x+3+x+x+3=170.

Simplifiyng, 4x+6=170

subtract six from both sides. Get 4x=164

Divide by 4. Get x=41

add three. Get length = 44

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Martha's weight is 75kg. Her weight increased by 3%. What is her new weight?
Brums [2.3K]
Martha’s new weight is 77.25. 3% of 75 is 2.25 and you can find that by multiplying 75 by .03 and adding that number to 75. hope this helps!
3 0
3 years ago
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15 POINTSSS ANSWER ASAP PLEASE!!!!! What is the equation, in point-slope form, of the line that is
Gnesinka [82]

Answer:

The Answer is : D

Step-by-step explanation:

First find the slope of the line that the equation should be parallel to. In this case it is 2/1 which simplified is 2.

Next insert the X (4) of the point (4,1) and solve to see if you get the Y(1).

y-1 = 2 (4-4)

y-1= 2 (0)

y-1= 0

y= 1

In this case D is correct.

TIP*

If you see the question ask you about a parallel formula, look at the slopes of them to see if they match up. Parallel formulas have the same slope, just a tip because you can see in the answer choices none of the equations have the same slope as the line on the graph except for D.

4 0
3 years ago
together Alice & Mindy have saved $490 for a trip this summer. Amy has saved $54 more than Mindy. How much has each person s
Rus_ich [418]

Answer:

Alice saved 218 and mindy saved 218+54

Step-by-step explanation:

4 0
3 years ago
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The average daily jail population in the United States is 706,242. If the distribution is normal and the standard deviation is 5
Karolina [17]

a. The probability that on a randomly selected day, the jail population

is greater than 750,000 is 20.1%

b. The probability that on a randomly selected day, the jail population is

between 600,000 and 700,000 is 43.2%

Step-by-step explanation:

The given is:

1. The average daily jail population in the United States is 706,242

2. The distribution is normal and the standard deviation is 52,145

3. We need to find the probability that on a randomly selected day,

    the jail population is greater than 750,000

4. We need to find the probability that on a randomly selected day,

    the jail population is between 600,000 and 700,000

a.

At first find z-score

∵ z = (x - μ)/σ, where x is the score, μ is the mean and σ is the standard

   deviation

∵  x = 750,000 , μ = 706,242 and σ = 52,145

∴ z = \frac{750,000-706,242}{52,145} ≅ 0.84

Use the normal distribution table of z to find the area to the right of

the z-value

∵ The corresponding area to z-score of 0.84 is 0.79955

- But we are interested in x > 750,000, we need the area to the

  right of z-score

∴ P(x > 750,000) = 1 - 0.79955 = 0.2005

∴ P(x > 750,000) = 0.2005 × 100% = 20.1%

The probability that on a randomly selected day, the jail population is

greater than 750,000 is 20.1%

b.

We will find z-score for 600,000 < x < 700,000

∵ z = \frac{600,000-706,242}{52,145} ≅ -2.04

∵ z = \frac{700,000-706,242}{52,145} ≅ -0.12

Use the normal distribution table of z to find the area between

the two z-values

∵ The corresponding area to z-score of -2.04 is 0.02068

∵ The corresponding area to z-score of -0.12 is 0.45224

- To find P(600,000 < x < 700,000) subtract the two values above

∴ P(600,000 < x < 700,000) = 0.45224 - 0.02068 = 0.4316

∴ P(600,000 < x < 700,000) = 0.4316 × 100% = 43.2%

The probability that on a randomly selected day, the jail population is

between 600,000 and 700,000 is 43.2%

Learn more:

You can learn more about z-score in brainly.com/question/7207785

#LearnwithBrainly

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%20-%206%3D%20-55" id="TexFormula1" title="x - 6= -55" alt="x - 6= -55" align="absmiddle" cla
Helga [31]

Answer:

<h2>x = -49</h2>

Step-by-step explanation:

x - 6 = -55      <em>add 6 to both sides</em>

x - 6 + 6 = -55 + 6

x = -49

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3 years ago
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