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2 years ago

What integer is the closest to 731

Mathematics

1 answer:

guajiro [1.7K]2 years ago

6 0

Answer:

731 is an integer. An Integer is a whole number!

Step-by-step explanation:

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Select all the correct answers.

Tanya [424]

Answer:

x=24

y=21

Step-by-step explanation:

ok budy mark me as brainlsit

8 0

3 years ago

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What is 8/21 in simplifeyd

telo118 [61]

The fraction 8/21 cannot be simplified because the numerator and denominator don't have a common factor except 1.

4 0

3 years ago

The mean for a set of 5 numbers is 24. Four of the numbers 18,30,24, and 27. What is the fifth number?

ioda

Answer:

21 it is the fifth number

5 0

2 years ago

ASAP PLEASE ANSWER

pickupchik [31]

Answer:

Step-by-step explanation:

Hope this helps!

7 0

3 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 $24.493$

3 0

3 years ago

Read 2 more answers