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3 years ago

What is the constant of proportionality, or unit rate, or the following table?

Mathematics

1 answer:

Alex787 [66]3 years ago

7 0

Wheres the picture...................................................

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A school has an enrollment of 600 students. 330 of the students are girls. Express the fraction of students who are boys to tota

boyakko [2]

Just do 100*600 then make the 330 into 400 and do 10 millameter and 10 inch from 11 inches and add all them up and call FEDEX to give you a box and your mail. I hope you understand

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3 years ago

Read 2 more answers

Write the slope-intercept form of the equation of each line. 9x-7y=-7

makkiz [27]

Slope intercept form is y = mx + b.....so lets solve for y

9x - 7y = -7 .....subtract 9x from both sides
-7y = -9x - 7...now we divide both sides by -7
y = 9/7x + 1 <==

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3 years ago

I need help on part b plz

Jet001 [13]

0.7, 0.8, 0.9. BAM !

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4 years ago

Jenna has 6 yards of yarn to knit sweaters for her 3 sisters. Each sweater uses 3⁄4 of a yard of yarn to make.

valentina_108 [34]

A. 8 sweaters
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3 years ago

A major department store chain is interested in estimating the average amount its credit card customers spent on their first vis

SSSSS [86.1K]

Answer: (39.424, 61.576)

Step-by-step explanation:

When population standard deviation( $\sigma$ ) unknown ,The confidence interval for population mean is given by :-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size

$\overline{x}$ = sample mean.

s= sample standard deviation

$t^*$ = Critical t-value (two-tailed)

Given : n= 15

Degree of freedom= 14 [df=n-1]

$\overline{x}=\ $50.50$

$s^2=400\\\\\Rightarrow\ s=\sqrt{400}=20$

Significance level = $\alpha=1-0.95=0.05$

For $\alpha=0.05$ and df = 14, the critical t-values : $t^*=\pm2.1448$

Then the 95% confidence interval for population mean will be :

$50.50\pm (2.1448)\dfrac{20}{\sqrt{15}}\\\\=50.50\pm(2.1448)(5.1640)\\\\\approx50.50\pm11.076\\\\=(50.50-11.076,\ 50.50+11.076)\\\\=(39.424,\ 61.576)$

Hence, a 95% confidence interval for the average amount its credit card customers spent on their first visit to the chain's new store in the mall. : (39.424, 61.576)

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4 years ago