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3 years ago

Select the slope-intercept form of an equation for a line with y-intercept -5 and slope 2.

Mathematics

2 answers:

Goryan [66]3 years ago

5 0

Answer:

y = 2x - 5

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

m - slope

b - y-intercept

xeze [42]3 years ago

3 0

Answer:

y = 2x - 5

Step-by-step explanation

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

Now, we have to substitute the values to the equation.

This will be our equation ---> y = 2x -5

Hope it helps!

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What is the slope of a line that is perpendicular to the line represented by the equation y=3/4x-1/2?

kherson [118]

The answer is: " - (4/3) " .
____________________________________
Explanation:
____________________________________
The original equation given has a slope of (3/4).

Note: We know this since the equation for the slope of the line is written in "slope-intercept form" ; also known as: "point-slope form"; that is:

" y = mx + b " ; in which "m" (the coefficient of "x") is the slope.
____________________________________________________
The slope of a line PERPENDICULAR to an equation, when written in "slope-intercept form", is the "negative reciprocal" of the slope of the original line.

Hence, the negative reciprocal of "(3/4)" is: "-(4/3)" .
____________________________________________________

5 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

When Zack checked his solution, it didn’t work. What was his mistake?

dangina [55]

Answer: its c In step 3, he should have divided both sides by StartFraction 6 Over 4 Endfraction.

Step-by-step explanation:

6 0

3 years ago

If x+a is a factor of 2x²+2ax+5x+10,find a.

vazorg [7]

If $x+a$ is a factor of the polynomial, it means that $-a$ is a root of the polynomial.