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3 years ago

7

Nita is using ribbon for gift boxes. She has 49 feet of ribbon. She needs to wrap 16 boxes of identical size. How much ribbon ca

n she use on each box?
3 feet

3 1/16 feet

3 1/8 feet

3 7/16 feet

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1 answer:

OLEGan [10]3 years ago

5 0

Answer:

The answer would be B or 3 1/16.

Step-by-step explanation:

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4(x-3)-(2x+5)=-3x-27

Georgia [21]

4(x-3)-(2x+5)=-3x-27
4*x - 4*3 - 2x - 5 = -3x - 27
4x - 12 - 2x - 5 = -3x - 27
2x - 17 = -3x - 27
2x + 3x = -27 + 17
5x = -10 / : 5
x = -2

7 0

2 years ago

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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

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Thelma and Louise drove 294 miles and used 14 gallons of gas. How many miles did their car get per gallon of gas?

Ann [662]

Divide 294 by 14 and you get 21. I hope it helps and that’s the answer. :)

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3 years ago

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I need help like quick plssss

Vera_Pavlovna [14]

Answer:

172.8

Step-by-step explanation:

24+18+5+3=50

360/50=7.2

7.2x24=172.8

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3 years ago

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PLS HELP ME I DON'T HAVE A LOT OF TIME !!!

Sergeu [11.5K]

Answer:

d

Step-by-step explanation:

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2 years ago