F(x) = x2 + 3; g(x) = Square root of quantity x minus two.<br><br> Find f(g(x)).

Maria is planning to attend UCLA. She is curious what the average age of UCLA students is. Since most students that attend UCLA

RoseWind [281]

Answer:

Yes

We fail to reject the alternative hypothesis Hₐ < 30, that is the average age of students is less than 30

Step-by-step explanation:

Yes because, there is an average age of a sample which can be tested against a null hypotheses

We put

Null hypothesis as H₀ = 30

The alternative hypothesis as Hₐ < 30

We have

$z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

With

$\bar x$ = 29

μ = 30

σ = 5.2

n = 65

α = 10%

We have

Here we have z = -1.55 and critical z = -1.28

Which gives a critical $\bar x$ of 29.83 with the probability P = 0.061 < 0.1 Hence we reject the null hypothesis as there is sufficient evidence to suggest that the average age is less than 30 years.

8 0

3 years ago

Please someone help I’m in the middle of my test and I don’t understand this question please someone help me answer it.

DENIUS [597]

Answer:

Freedom of the American people

broadened as a result of the Civil War.Freedom of the American people

broadened as a result of the Civil War.

Step-by-step explanation:

8 0

2 years ago

Sue deposited $1,500 into two different accounts.

avanturin [10]

$915 I’m pretty sure

5 0

3 years ago

Solve the system of equations

ikadub [295]

Answer:

The answer is B.

Step-by-step explanation:

4 0

3 years ago

Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2

Inessa05 [86]

Answer:

$f_{ave}$ = ${\frac{10 }{3\pi }$

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3 $\pi$ /2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

$f_{ave}$ = $\frac{1}{b - a}\int\limits^b_a {f(x)} \, dx$

Now,

Here a = 0, b = $\frac{3\pi }{2}$ , f(x) = $5sec^{2}(\frac{x}{6} )$

Now,

$f_{ave}$ = $\frac{1}{\frac{3\pi }{2} - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ]$

= ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]$

= ${\frac{10 }{3\pi }}[ {1 - 0 } ]$

= ${\frac{10 }{3\pi }$

⇒ $f_{ave}$ = ${\frac{10 }{3\pi }$

3 0

2 years ago

F(x) = x2 + 3; g(x) = Square root of quantity x minus two. Find f(g(x)).