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3 years ago

Help me 9/16 / 7/10 =

Mathematics

2 answers:

lbvjy [14]3 years ago

6 0

45/56 because 9/16 \ 7/10 = 9/16 x 10/7 = 90/112 which is 45/56 in simplest terms

Sholpan [36]3 years ago

4 0

45/56 is your answer

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Assume the population of human body temperatures has a mean of 98.6 degrees Fahrenheit as is commonly believed. Assume the stand

Dennis_Churaev [7]

The probability that the mean of a sample of 106 randomly selected humans is lower than 98.5°F is 4.85%

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Z score is given as:

z = (raw score - mean) ÷ (standard deviation/√sample size)

Given mean of 98.6°F, standard deviation is 0.62°F, sample size = 106

For x < 98.5:

z = (98.5 - 98.6) ÷ (0.62÷√106) = -1.66

P(z < -1.66) = 0.0485

The probability that the mean of a sample of 106 randomly selected humans is lower than 98.5°F is 4.85%

Find out more on equation at: brainly.com/question/2972832

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4 0

2 years ago

A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th

Burka [1]

Answer:

Confidence interval : $21.506$ to $24.493$

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = $\bar{x}=\frac{\sum x}{n}$

Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$

Sample standard deviation= s = $1.72$

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table $t_{\frac{\alpha}{2}$ = 3.883

Formula of confidence interval : $\bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}$

Substitute the values in the formula

Confidence interval : $23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $23 -3.883 \times \frac{1.72}{\sqrt{20}}$ to $23 + 3.883 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $21.506$ to $24.493$

Hence Confidence interval : $21.506$ to $24.493$

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