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3 years ago

X/w = z/y². solve for y

Mathematics

1 answer:

jek_recluse [69]3 years ago

8 0

Answer:

Step-by-step explanation:

We can simplify (cross multiply) to get $xy^2=wz$

Then we easily simplify to get $y=\sqrt{\frac{wz}{x}},\:y=-\sqrt{\frac{wz}{x}};\quad \:w\ne \:0,\:x\ne \:0$

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Simora [160]

Answer:

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Step-by-step explanation:

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2 years ago

Combine 2y+1x=40 and y=2x using substitution

fomenos

2(2x) + 1x = 40 or 4x + 1x = 40 is the result of combining by substitution

Solution:

Given that we have to combine 2y + 1x = 40 and y = 2x using substitution method

The substitution method for solving systems of equations involves expressing one variable in terms of another, thus removing one variable from an equation.

Given equations are:

2y + 1x = 40 -------- eqn 1

y = 2x ----------- eqn 2

We can substitute eqn 2 in eqn 1

Which means, substitute y = 2x in place of y in eqn 1

2(2x) + 1x = 40

4x + 1x = 40

5x = 40

x = 8

From eqn 2,

y = 2(8)

y = 16

Thus by combining using substitution method we found the solution

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3 years ago

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Andrew [12]

Answer:

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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

Write an equation that uses partial products to multiply 8 X 64

zhuklara [117]

8 x 64 = (8 x 60) + (8 x 4)=

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3 years ago

Read 2 more answers

X/w = z/y². solve for y​

X/w = z/y². solve for y