Y=4x+2 passes through point (3,-5)

ASAP! Question 3: The following data give the weights (in pounds) lost by 15 members of a health club at the end of 2 months aft

Klio2033 [76]

Answer:

First arrange the data in an order

3, 3, 3, 7, 7, 10, 10, 12, 12, 13, 14, 17, 17, 18, 21

a.) Q1 = (N + 1)/4th item = (15 + 1)/4 = 4th item = 7

Q2 = 2(16)/4 = 16/2 = 8th item = 12

Q3 = 3(16)/4 = 3(4) = 12th item = 17

IQR = (Q3 - Q1)/2 = (17 - 7)/2 = 10/2 = 5

b.) 82nd percentile = 82(16)/100 = 13th item = 17

c.) 12 is the 8th item which is in the middle of the data set, so the percentile rank is 50%

Step-by-step explanation:

3 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

Help on Math please and thank you

Keith_Richards [23]

Answer:

the value of x to the nearest tenth is 12 * x

5 0

3 years ago

Does anyone know how to do this please help me

Anit [1.1K]

Answer:

162°

Step-by-step explanation:

a circle has 360°, or the angle of a whole circle is that

so 45% of 360 =

0.45 times 360 = 162

7 0

3 years ago

A hardware dealer has 168kg of nails, if one nail weighs 0.25kg, how many nails he have?

snow_tiger [21]

Answer:

He has 672 nails

Step-by-step explanation:

168/.25

7 0

3 years ago

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