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If the length of a rectangular field is twice the width and the perimeter is 24. Write an equation that an be used to solve for

Sedbober [7]

If we use the width as x we have 2x with both widths and 4x with both lengths. So 6x=24 and x=4. So length is 8 and width is 4.

7 0

4 years ago

Free

Leona [35]

Answer:

thanks appreciated it

3 0

3 years ago

Read 2 more answers

You have 4 cards, 2 black and 2 red. You play a game where during each round you draw a card. If it's black, you lose a point. I

sineoko [7]

Answer: The expected value of this game is 2/3

Step-by-step explanation:

Give that

If it's black, you lose a point. If it's red, you gain a point.

And then you can stop at any time. But you should never stop when you are losing because that can guarantee 0 by drawing all the cards.

Assuming you should stop after three cards when you are +2.

The only question is whether to draw if you are +1 on the first draw.

If you draw red first, You have 1/3 chance of drawing red again and this will give you +2 points

1/3 chance of drawing two blacks and earn zero point, chance of drawing black-red and earn +1. This gives +1, so it doesn't matter whether you draw or not.

From the beginning, If you draw red (probability 1/2 you end +1. If you draw black and then draw two reds (probability 1/6 you end +1) Otherwise you break even with probability 1/3. Overall, the value is 2/3

8 0

3 years ago

Solve the system of equations by graphing on your own paper. Write the

Verdich [7]

Answer:

(9, -1)

Step-by-step explanation:

7 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

4 years ago