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4 years ago

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.

First, we need to structure the total coupon formula:

Total coupon= 5*x

x= pack of four books

Now, for $15

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 8x+4=6x-10 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 standard deviation for the sample, 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