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Doss [256]
3 years ago
8

For the function f(x)=5x-3 and g(x)=x+4 which composition produces the greatest output

Mathematics
2 answers:
mr_godi [17]3 years ago
4 0
The first one: 5(x+4)-3=5x+20-3=5x+17
the second one: 5x-3+4=5x+1
when you substitute 1 with both the functions: 5*1+17=22 and 5*1+1=6
the answer would be: (A) produces the greatest output. i think that is how you do it.
Dima020 [189]3 years ago
3 0
F(g(x)) = 5(x+4)-3
            = 5x+20-3
            = 5x+17

G(f(x)) = 5x-3+4
           = 5x+1

Therefore, f composed of g produces the greatest output.

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Answer:

-3.16666666667

Step-by-step explanation:

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HELP NEEDED I need some help here can anyone help please?
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Suppose the average of 15 consectutive numbers is 15. What is the average of the first five numbers of the set?
Marina86 [1]
15x15= 225

15x + 105= 225

15x= 225-105

15x= 120

X=8

X+ (x+1) + (x+2) + (x+3)…etc all the way up too +(x+14)=225

The first five numbers of the set: 8, 9, 10, 11, 12
7 0
1 year ago
Which expression is equivalent to this expression? <br><br> thank you :)
CaHeK987 [17]

Answer:

C. -4x - 1

Step-by-step explanation:

-\frac{1}{2}(8x+2)

Distribute the term outside of the parenthesis, Multiply each term in the parenthesis by the term outside (do not forget to include the negative sign when multiplying);

(-\frac{1}{2}(8x))+(-\frac{1}{2}(2))

Simplify,

-4x - 1

6 0
2 years ago
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
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