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Maurinko [17]
2 years ago
13

PLEASE HELP

Mathematics
2 answers:
tekilochka [14]2 years ago
7 0

Answer:

PLEASE HELP

For the right triangles below, find the values of the side lengths b and c.

Round your answers to the nearest tenth.

Step-by-step explanation:

PLEASE HELP

For the right triangles below, find the values of the side lengths b and c.

Round your answers to the nearest tenth.

Damm [24]2 years ago
6 0

Answer:

b = 1.7

c = 7.1

Step-by-step explanation:

the triangle with side 'b' is a 30-60-90° Δ where the ratio of sides, respectively, are 1 : \sqrt{3} : 2

to find 'b' we could set up this proportion:

b/3 = 1/\sqrt{3}

cross-multiply:

b\sqrt{3} = 3

b = 3 / \sqrt{3}

if we rationalize the denominator, this becomes (3\sqrt{3})/3 which is \sqrt{3} or 1.7

the other triangle is a 45-45-90°Δ where the ratio of the sides are 1 : 1 : \sqrt{2}

if the legs are 5 then the hypotenuse is 5\sqrt{2} or 7.1

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Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

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