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BlackZzzverrR [31]
2 years ago
6

When solving this equation for x, what would you do next?

Mathematics
1 answer:
Leviafan [203]2 years ago
4 0

Answer: A. Distribute the 3 outside of the parathesis to everything inside the parenthesis

We use the order of operations, also known as PEMDAS.

Since there is a variable inside the parenthesis, we cannot add/subtract 4. This means the next step is to distribute.

After distributing, the equation looks like this and we can solve it from there:

4x + 12 = 30

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In the figure, the perimeter of hexagon ABCDEF is approximately
Nikitich [7]
First solve the length of side BC, CD, EF and FA   Since BC = CD = sqrt( 10^2 + 10^2) BC = CD = 14.1421  
FA = EF = sqrt(10^2 + 20^2) = 23.3607  
So the perimeter = 10 + 10 + 14.1421 + 14.1421 + 23.3607 = 93

The area is made up be triangle FAE, rectangle ABDE and triangle BCD
A = 0.5(20)(20) + (10)(20) + 0.5(20)(10)
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3 years ago
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In 2016, a town had a population of 38,500. With a population growth of 3.4%, what would
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3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
3 years ago
in my numbers l. the digit in the ones place is double the digit in the tens place. the sum of the digits is 3. my number is?
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is it 1 + 2=3. are is it 2+1=3. are you a  boy  are  not

6 0
3 years ago
The area of a rectangle is 63m squared,and the length of the rectangle is 5 m more than twice the width.Find the dimensions of t
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Given:
Area(A)= 63m^2
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Width (W)=?

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0=(2W-9)(W+7)
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Can only use 4.5 since it is positive and distance is positive.
W= 4.5 m
L=2W+5=2 (4.5)+5=9+5=14m

7 0
3 years ago
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