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kherson [118]
3 years ago
8

A car travels 172 miles using 7 gallons of gas. At that rate, how far can the car traval using 42 gallons of gas?

Mathematics
1 answer:
Semmy [17]3 years ago
7 0

Answer:

1.1032

2.6

3.16

4.35:50 I think

Step-by-step explanation:


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Erick paid a $75 registration fee to join the gym. He pays $15 each month.
liraira [26]

Answer:

$150

Step-by-step explanation:

Erick has to pay $15 per month so for 5 months he will have to pay $75 because 15 times 5 is 75. So $75 for registration fee + $75 for 5 months to use the gym give us $150

7 0
3 years ago
CAN SOMEONE HELP MEE
Artist 52 [7]

Answer:

11.45

Step-by-step explanation:

the circumference is 36 so divide it by pie then rounded it to the nearest hundredths place

7 0
3 years ago
The width of a rectangle measures (5v-10w)centimeters, and its length measures (2v+3w)centimeters. Which expression represents t
kiruha [24]

Answer:the perimeter of the rectangle =A) (−14w+14v) centimeters

Step-by-step explanation:

Step 1

The perimeter of a Rectangle is given as 2 (l+ w)

If Length = 2v+3w centimeters.

and width =5v-10wcentimeters

Then Perimeter =2 (2v+3w + 5v-10w)

Perimeter=2(7v -7w)

Perimeter= 14v - 14w or can be written as -14w +14v

8 0
3 years ago
Select the expression that means y divided by 3.
LenKa [72]

Answer:

E. y/3

Step-by-step explanation:

Read and follow:

Y DIVIDED BY 3

Y         ÷            3

or just

\frac{y}{3}

Hope this helps

~R3VO

5 0
3 years ago
Read 2 more answers
In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden
hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is \hat{p}=26/42=0.6190. The estimated standard deviation is given by \sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749. Because we have a large sample, the 90% confidence interval for p is given by 0.6190\pm z_{0.05}0.0749 where z_{0.05}=1.6448 is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is 0.6190\pm (1.6448)(0.0749), i.e., (0.4958, 0.7422).

5 0
3 years ago
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