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-BARSIC- [3]
3 years ago
14

If f(x) =5x-25 and g(x) = (1/5)X+5, which expression could be used to verify g(x) is the inverse of f(x)

Mathematics
2 answers:
mina [271]3 years ago
7 0
Answer:
f(g(x)) = x

Explanation:
In order to prove that one function is the inverse of the other, all you have to do is substitute in the main function with the inverse one and solve. If the result is x, then it is verified that one function is the inverse of the other. 
Now for the given functions we have:
<span>f(x) =5x-25
</span><span>g(x) = (1/5)x+5
We want to prove that g(x) is the inverse of f(x).
Substitute in the above formula and compute the result as follows:
f(g(x)) = 5(</span>(1/5)x+5) - 25
          = x + 25 - 25
          = x
The final result is "x", therefore, it is verified that g(x) is the inverse of f(x)

Hope this helps :)
Elanso [62]3 years ago
6 0
If g(x) is the inverse of f(x) the  f(g(x)) = x

f(g(x)) =  5 ( 1/5x + 5) - 25 

if you simplify this you'll get x  so it is the inverse.
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x=32

Step-by-step explanation:

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True or False: Only like terms can be combined. A True B False
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Answer:

True

Step-by-step explanation:

Like terms like 4x-3x the x term. is a like term

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4 0
3 years ago
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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in
Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0
2 years ago
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