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Dmitrij [34]
3 years ago
8

Can someone help me with this question?

Mathematics
2 answers:
Nastasia [14]3 years ago
5 0
To find out whether n = 112 is a solution to the inequality, plug 112 for n into the inequality and simplify.

10 + n/28 > 14
10 + 112/28 > 14
10 + 4 > 14
14 > 14

This reads "fourteen is greater than fourteen," which is false.

n = 112 is NOT a solution to the inequality 10 + n/28 > 14.
Akimi4 [234]3 years ago
3 0
Hey there! :D

So, 10+112/28=14 
So 14 > 14
It's not true.
So it is NO.
Hoping this helps you!

Have a good day my friend!!:D
My pleasure :)
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Explain how to factor when the leading coefficient is 1.
Bas_tet [7]

Answer:  C) Find the factors of c that add up to b.

==============================================

Explanation:

If we want to factor something in the form x^2+bx+c, then we look for two numbers that

  • Multiply to c
  • Add to b

-------------

Let's look at a specific example

Consider factoring x^2+5x+6

We need to find two numbers that...

  • Multiply to c = 6
  • Add to b = 5

Through trial and error, you should find the two numbers to be 3 and 2. This means it factors to (x+3)(x+2). The order of the factors doesn't matter.

You can use the FOIL rule or the box method to expand out (x+3)(x+2). You should get x^2+5x+6 again.

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2 years ago
Rearrange the equation a = 1/2 b to make b the formula
Evgesh-ka [11]

Answer: b = 2a

Step-by-step explanation:

<u>Original equation</u>

a = (1/2) b

<u>Multiply 2 on both sides</u>

a * 2 = (1/2) b * 2

\Large\boxed{b=2a}

Hope this helps!! :)

Please let me know if you have any quesitons

8 0
2 years ago
Read 2 more answers
Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

Where (from tables):

Z_{0.99}=2.33

Finally, the interval at 98% confidence level is:

CI(\mu)=\lbrack28.94,31.06\rbrack

4 0
1 year ago
The horsepower , H(s), required for a racecar to overcome wind resistance is given by the function : H(s) = 0.003s^2+0.07s-0.027
mote1985 [20]
Average rate of change = [H(100) - H(80)] / (100 - 80)
H(100) = 0.003(100)^2 + 0.07(100) - 0.027 = 0.003(10000) + 0.07(100) - 0.027 = 30 + 7 - 0.027 = 36.973

H(80) = 0.003(80)^2 + 0.07(80) - 0.027 = 0.003(6400) + 0.07(80) - 0.027 = 19.2 + 5.6 - 0.027 = 24.773

Average rate of change = (36.973 - 24.773)/(100 - 80) = 12.2/20 = 0.61

Answer: B
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3 years ago
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Answer:

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Step-by-step explanation:

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