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Natasha_Volkova [10]

3 years ago

13

Find the length of the leg of a right triangle with leg length b= 21.5 inches and the hypotenuse c= 31.9 inches. Use a calculato

r to estimate the square root to one decimal place.

1 answer:

sweet [91]3 years ago

8 0

Answer:

23.6 (approximate value)

Step-by-step explanation:

a*a+b*b=c*c Pythagorean thereom
(21.5)^2 + b*b = (31.9)^2
b*b= 1017.61 - 462.25
√b*b= √555.36
b=23.6(approximate value)

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Degger [83]

Answer:

Male=21 Female=14

Step-by-step explanation:

So the problem would be (f)+(f+7)=35

first you would subtract 7 from 35

Now you have (f)+(f)=28 or 2f=28

Then you divide both sides by two

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and since we know there is 7 more males then the answer is

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3 years ago

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jerry deposits 80.00 into new savings account the account earns 3.5% simple interest per year no money is added or removed from

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3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

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P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago

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Nataly_w [17]

Answer:

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Step-by-step explanation:

that is the answer....

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3 years ago

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7 0

3 years ago