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Maru [420]
3 years ago
7

Need help with simplifying expressions

Mathematics
2 answers:
dolphi86 [110]3 years ago
5 0
The answer is -20

I know it looks hard, but all you need to do is combine like terms.

Here’s an example:

4x+5x=18

4x plus 5x equals 9x. So your equation is now-

9x=18

Now you just divide 18 by 9

which is 2, and you just drop the x

Hope this helps
Ilia_Sergeevich [38]3 years ago
3 0

Answer:

A) -20

Step-by-step explanation:

distribute the 4y and the -4 to the parentheses

16y^2 -16y+16y-20-16y^2

combine like terms

16y^2-16y^2=0

-16y+16y=0

-20

Mark brainliest??

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Step-by-step explanation: (dont trust me)

x^2 + (7)x +18

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25, 8, 10, 35, 5, 45, 40, 30, 20 What is the upper quartile of the data set above?A.35
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Katie is getting her laundry done it cost $2.25 per square foot of flowers added to the yard if she wants 12.4 ft.² of flowers h
7nadin3 [17]

Answer:

Katie needs to pay $27.9.

Step-by-step explanation:

Given;

Cost of per foot = $2.25

Number of  flowers in her yard = 12.4 square foot

We need to find how much she needs to pay.

Solution:

Now we know that;

1\ ft^2 = \$2.25

So for 12.4 ft^2 = Cost for 12.4 ft^2 .

By using Unitary method we get;

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Hence Katie needs to pay $27.9.

5 0
3 years ago
A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

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

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

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

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

<u><em>Final answer:-</em></u>

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0
3 years ago
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