Please help marking u Brainiest if answered

The hypotenuse of a right triangle is 6 inches and one of the legs is √ 6 inches, the exact value of the other leg is _______ in

KonstantinChe [14]

The answer is 3.55 inches

3 0

3 years ago

16. Suppose Eva has 12 3/4 pounds of flour. The day's baking will require triple that amount.

egoroff_w [7]

Answer:

Option D

Step-by-step explanation:

Given the following question:

$12 \frac{3}{4} \times3$

In order to find the answer, we have to convert the mixed number and then multiply by three, since how Eva needs triple the amount for today's baking.

$12 \frac{3}{4} \times3$
$12\frac{3}{4} =4\times12=48+3=51=\frac{51}{4}$
$3=\frac{3}{1}$
$\frac{51}{4} \times\frac{3}{1}$
$51\times3=153$
$4\times1=4$
$=\frac{153}{4}$

Convert the improper fraction back into a mixed number:

$\frac{153}{4}$
$\frac{153}{4} =153\div4=38\frac{1}{4}$
$=38\frac{1}{4}$

Which means Eva needs "38 1/4 pounds of flour" or option D for today's baking.

3 0

3 years ago

How do you write 7,020,032 in expanded forms.

yulyashka [42]

7,000,000. 20,000. 30. 2

3 0

4 years ago

Calculate the unit price of the following item:

Vikki [24]

10 lol that’s the answer.

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