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2 years ago

1. Six units to the left of 3

Mathematics

1 answer:

trapecia [35]2 years ago

8 0

Answer:

six uniys to left of 3 is= +2,+1,0,-1-2-3

three unit right of 15is= +16,+17, +18

Eight unit to the left of =-4,-5,-6,-8,-9,-10,-11,-12

seven units ti right of -5=-3,-2,-1,0+1,+2,+3,+4

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Montano1993 [528]

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1 year ago

You have 400 cards stored 3/5 of them are from family. 1/4 of them are from friends. The rest you made. How many cards did you m

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3 years ago

Which of the following is a perfect square trinomial?

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$D.~4x^2-20xy+25y^2$

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2 years ago

Solve the inequality y-12<-5

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2 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.

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3 years ago