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nevsk [136]
3 years ago
15

The area of a rectangle is 42 square millimeters. The length is 7 millimeters. What is the perimeter of the rectangle?

Mathematics
1 answer:
Aleks [24]3 years ago
8 0

Answer:

26 mm

Step-by-step explanation:

A=wl= 42 mm²

l=7, w= 42/7= 6 mm

P=2(l+w)= 2(7+6)= 26 mm

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4a - b = 3, solve for a​
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Answer: your answer is

7a hope this helps you

Step-by-step explanation:

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The radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles. a. Find the surface area of Earth and the m
Molodets [167]

A) The formula for surface area of a sphere is A = 4*PI*r^2

using 3.14 for PI:

Surface area for Earth = 4 * 3.14 x 3960^2 = 196,960,896 miles^2

Surface area of the moon: 4 * 3.14 * 1080^2 = 14,649,984 miles^2

B)Divide the Surface of the Earth by the moon:

196,960,896 / 14,649,984 = 13.44

The Earths surface is 13.4 times larger than the moon.

C) Multiply the surface of the Earth by 70%:

196,960,896 * 0.70 = 137,872,627.2 million square miles of water.

4 0
3 years ago
What is the equation for standard error of the mean?
Flauer [41]
When your ball drops on the floor
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3 years ago
A manufacturing process makes rods that vary slightly in length but follow a normal distribution with mean length 25 cm and stan
Galina-37 [17]

Answer:

The probability of randomly selecting a rod that is shorter than 22 cm

P(X<22)  = 0.1251

Step-by-step explanation:

<u><em>Step(i):</em></u>-

Given mean of the Population = 25cm

Given standard deviation of the Population = 2.60

Let 'x' be the random variable in normal distribution

Given x=22

Z = \frac{x-mean }{S.D} = \frac{22-25}{2.60} = -1.15

<u><em>Step(ii):</em></u>-

The probability of randomly selecting a rod that is shorter than 22 cm

P(X<22) = P( Z<-1.15)

             = 1-P(Z>1.15)

             =  1-( 0.5+A(1.15)

             =  0.5 - A(1.15)

             = 0.5 - 0.3749

             = 0.1251

The probability of randomly selecting a rod that is shorter than 22 cm

P(X<22)  = 0.1251

7 0
3 years ago
I need help with number 33!!!
sveta [45]

Area of Big Rectangle                                     Area of Small Rectangle

A = L x w                                                           A = L x w

 = (25 + 2w)(10 + 2w)                                           = 25 x 10

 =   250 + 50w + 20w + 4w²                               = 250

       Big - Small  

74 = 250 + 70w + 4w² - 250

74 = 4w² + 70w

0 = 4w² + 70w - 74

   = 2(2w² + 35w - 37)

  = 2[2w² -2w  |  +37w - 37]

  = 2[2w(w - 1) |  +1(w - 1)]

  = 2(2w + 1)(w - 1)

0 = 2                                   0 = 2w + 1                                     0 = w - 1

FALSE so disregard          w = -1/2                                         w = 1

                                         NEGATIVE so disregard

Answer: 1 meter

5 0
3 years ago
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