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Drupady [299]
4 years ago
10

1/10 is greater than??

Mathematics
1 answer:
Shtirlitz [24]4 years ago
5 0
\frac{1}{10}  \geq  \frac{1}{12} Think of when your mom cuts a pizza. 
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At a certain bookstore, you get a $5 coupon for every 4 books you buy. What is the least number of books you could buy to get $1
ira [324]

Answer:

You need to buy 12 books to get a $15 coupon.

Step-by-step explanation:

Giving the following information:

You get a $5 coupon for every 4 books you buy.

<u>First, we need to structure the total coupon formula:</u>

Total coupon= 5*x

x= pack of four books

<u>Now, for $15</u>

15 = 5x

15/5= x

3= x

4*3= 12

You need to buy 12 books to get a $15 coupon.

6 0
3 years ago
A random sample of 15 of the 78 competitors at a middle school gymnastics competition are asked
bija089 [108]

Answer: 60

Step-by-step explanation:

mean: you just add all 15 numbers then divide the number you get by 15.

55+57+57+58+59+59+59+59+59+61+62+62+63+64+66=900

900/15=60

5 0
3 years ago
PLEASE HELP ME I REALLY NEED HELP NOT GOOD AT AREA PLEASE HURRY
Gnoma [55]
1: 132in^2

2: 275units^2

3: 139.5in^2

4: 168ft^2

5: 228ft^2
8 0
3 years ago
Solve for x<br> 8x+4=6x-10<br> x=?
mamaluj [8]

8x+4 = 6x-10

To solve for x you'll want to move all the terms containing x onto the left side of the equation and all the other terms on the right side of the equation.

1. Subtract 6x from both sides.

2x+4 = -10

2. Subtract 4 from both sides.

2x = -14

3. Divide both sides by 2.

x = -7

Answer is:

-7

7 0
3 years ago
Read 2 more answers
A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups
mash [69]

Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).

For this problem, we have the <u>standard deviation for the sample</u>, thus, the t-distribution is used.

  • The sample mean is of 15, thus \overline{x} = 15.
  • The sample standard deviation is of 9, thus s = 9.
  • The sample size is of 10, thus n = 10.

First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.

Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622.

The margin of error is of:

M = t\frac{s}{n}

Then:

M = 2.2622\frac{9}{\sqrt{10}} = 6

The confidence interval is:

\overline{x} \pm M

Then

\overline{x} - M = 15 - 6 = 9

\overline{x} + M = 15 + 6 = 21

The confidence interval is (9, 21).

A similar problem is given at brainly.com/question/25157574

6 0
3 years ago
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