Find the highest common factor of 12 16 and 20

Given the functions f(x) = 7x + 13 and g(x) = x + 2, which of the following functions represents f[g(x)] correctly? (2 points)

Svetach [21]

Answer:

The value of f[ g(x) ] = 7x + 27

Step-by-step explanation:

It is given that, f( x ) = 7x + 13 and g( x ) = x + 2

To find the value of f(g(x))

g(x) = x + 2 and 7x + 13 (given)

Let g(x) = x + 2

f [ g(x) ] = 7(x + 2) + 13 [ substitute the value of g(x) in f(x) ]

= 7x + 14 + 13

= 7x + 27

Therefore the value of f[ g(x) ] = 7x + 27

8 0

3 years ago

7t+2r-3t+r what is the answer

Troyanec [42]

This simplifies to 4t+3r

8 0

3 years ago

Read 2 more answers

4x-5,4(x-5),4(x+5) drag the phrases into each box to match each expression five sang the sum of four and number five more than t

jeyben [28]

Answer:

4x-5= 5 less than the product of 4 and a number. 4 (x-5)= The product of 4 and the difference of a number and 5. 4(x+5) = The product of 4 and the sum of a number and 5.

Step-by-step explanation:

I took the test online in K12 and when I reviewed it said that those are the correct answer.

8 0

3 years ago

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

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 strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

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 $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

1 4/5 cm by 4 2/5 cm

LenaWriter [7]

If it is area you're looking for, the answer is 7.92 cm^2, or ~7 11/12 cm^2.

7 0

3 years ago

Find the highest common factor of 12 16 and 20​

Find the highest common factor of 12 16 and 20