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Deffense [45]
2 years ago
11

Find (gof)(3) f(x) = |x+2| g(x) = -x^2 A. -25 B. 1 C. 4 D. -15

Mathematics
2 answers:
aleksklad [387]2 years ago
5 0

First of all, we replace the unknown "x" by "3".

We put the function "f" inside the function "g".

g( |3 \:  +  \: 2| ) \:  =  \:  - |3 \:  +  \: 2| ^{2}

g( |5| ) \:  =  \:  -  |5|  ^{2}

g(5) \:  =  \:  - 5 ^{2}

g(5) \:  =  \:  \boxed{ \bold{ - 25}}

<h2>Answer: </h2>

\text{A. }  \: \boxed{ \bold{-25}}

<h3><em><u>MissSpanish</u></em> </h3>

Ksenya-84 [330]2 years ago
3 0

Answer:

A

Step-by-step explanation:

to find (g ○ f)(3) , evaluate f(3) and substitute the value obtained into g(x)

f(3) = | 3 + 2 | = 5 , then

g(5) = - 5² = - 25

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The slope-intercept form:

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b - y-intercept

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6=4(1)+b

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3 years ago
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

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3 years ago
In a lilac paint mixture , 40% of the. Mixture is white paint ,20%is blue and the rest red how many cups of red paint are used
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