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3 years ago

Evaluate the given expression of m = 6. n = 12. 4m-17+2n

Mathematics

1 answer:

Dominik [7]3 years ago

4 0

First, substitute 6 in for m and 12 in for n. 4(6)-17+2(12)

Then, multiply the numbers in parentheses by the outer numbers. 24-17+24

Finally, add and subtract from left to right. 24-17=5, and 5+24=29

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Brut [27]

Answer:

$20

Step-by-step explanation:

Simple, if he sold 16 pretzels, and each one costs $1.25, then multiply the cost by how many:

16X1.25=20

So he got $20

I hope this helps!

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3 years ago

Triangle Review Last one (Math)

Anni [7]

Answer:

A=30

Step-by-step explanation:

Set your equation up as following:

180=90+(x+37)+(x+67)

180=90+37+67+2x

180-90-37-67=2x

-14=2x

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3 years ago

Read 2 more answers

Given f(x)=x3 and g(x)= 1-5x2, fine (fog)(x) and it’s Domain

andre [41]

Answer:

Option B. $f(g(x)) = (1-5x ^ 2) ^ 3$ all real numbers

Step-by-step explanation:

We have

$f(x) = x ^ 3$ and $g(x) = 1-5x ^ 2$

They ask us to find

(fog)(x) and it's Domain

To solve this problem we must introduce the function g(x) within the function f(x)

That is, we must do f(g(x)).

So, we have:

$f(x) = x ^ 3$

$g(x) = 1-5x ^ 2$

Then:

$f(g(x)) = (1-5x ^ 2) ^ 3$

The domain of the function f(g(x)) is the range of the function $g(x) = 1-5x ^ 2$ .

Since the domain and range of g(x) are all real numbers then the domain of f(g(x)) are all real numbers

Therefore the correct answer is the option b: $f(g(x)) = (1-5x ^ 2) ^ 3$

And his domain is all real.

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3 years ago

Read 2 more answers

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

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3 years ago

Please help another question

butalik [34]

Answer is D) Y. It's matching with the results given

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