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3 years ago

14

you buy a ring box as a birthday gift that is in the shape of a triangular prism. what is the least amount of wrapping paper nee

ded to wrap the box?

1 answer:

garri49 [273]3 years ago

6 0

What are the measurements

You might be interested in

Solve for X <br><br> Will give brainiest

ELEN [110]

Answer:

x = -3

Step-by-step explanation:

55 + x + 74 + 54 = 180

x + 183 = 180

x = -3

8 0

3 years ago

Pre-calc, Review the table of values for function h(x). (image attached)

GalinKa [24]

Answer:

$\lim_{x \to 10^+} h(x) = 18.5$

Step-by-step explanation:

As x approaches 10 from the right side, h(x) approaches 18.5 but never touches it.

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

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

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

<u><em>95% confidence interval for</em></u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [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].

<em>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.</em>

8 0

3 years ago

F(x)=3/5x+1 find f(-15)

Luda [366]

Nothing hahahahahah

3 0

3 years ago

the total cost after tax to repair Kimber’s cracked phone is represented by 00.4(30h)+30h Where h represents the number of hours

Serjik [45]

we have that

the total cost is

$0.040(30h)+30h$

In this equation the term 30h represent the cost due to h hours at at unit of 30

and the term 0.040(30h) is equal to calculate the 4% of (30h)

so

in this problem the tax rate is equal to 4%

therefore

<u>the answer is</u>

The term 0.040(30h) represents the amount of tax she must pay

7 0

3 years ago