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3 years ago

Lou cuts 3 yards from a 9 yard roll of fabric. Then he cuts 4 feet from the roll. How many feet of fabric is left on the roll.

1 answer:

5 0

16 feet of fabric left on the roll. take 9 yards turn it into feet (times three to get feet) then do the same for 2 1/3. 9 yards becomes 27 feet and 2 1/3 yards becomes 7 feet. add 7 feet plus four. you get positive eleven. Then take 27 feet minus 11, you get 16 feet or 5 1/3 yards of fabric

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Rewrite without absolute value sign for different values of x: y=− 2x+5 + 2x−5 if x<−2.5; if x>2.5; if −2.5≤x≤2.5

valentina_108 [34]

Answer:

x<-2.5 =(2x+5)-(2x-5) 2. 5-5

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Step-by-step explanation:

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3 years ago

On a motor trip, Mr. Williams covered 2/6 of his total trip distance by driving 388 miles. The total distance to be covered in m

Reptile [31]

Answer:

Step-by-step explanation:

2/6 x d = 388

d = 388 x 6/2 = 1164 m

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2 years ago

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How do you do this problem? Explain.

Elza [17]

Answer:

The charges will be the same after 4 hours.

Step-by-step explanation:

Total Amount = y

Number of hours = x for x > 2

Garage A: y = $7.00 + (x - 2)*3

Garage B: y = 3.25*x

Part 3: What is the cost to be equal?

3.25x = 7 + 3(x - 2) Remove the brackets

3.25x = 7 + 3x - 6 Collect terms on the right

3.25x = 3x + 1 Subtract 3x from both sides.

3.25x - 3x = 3x - 3x + 1 Combine

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3 years ago

For science class, a student recorded the high and low temperatures, in Fahrenheit, over a ten day period in September. The data

Anestetic [448]

Thee correlation factor formula for two variables is
(r) =[ nΣxy – (Σx)(Σy) / Sqrt([nΣx^2 – (Σx)^2][nΣy^2 – (Σy)^2])].

A). as per the formula the correlation factor for the given data is 0.94.

B). the month of September has the highest correlation as 0.94 is much closer to 1 than 0.89.( note: 1 denotes the perfect correlation and 0 denotes no correlation at all)

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3 years ago

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A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon

fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that $n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694$

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

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2 years ago