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3 years ago

If you have $20 and spend $12 on food, what fraction of your money do you still have?

Mathematics

2 answers:

Bas_tet [7]3 years ago

8 0

Answer:

2/5

Step-by-step explanation:

please mark brainliest

just olya [345]3 years ago

6 0

Answer:

The fraction of the money that would be left would be 2/5. Hope this helps. :D

Step-by-step explanation:

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Please solve this for 15 points explain please

Likurg_2 [28]

Answer:

1.25 cups

Step-by-step explanation:

Since you have 5 5/8 cups = 4 1/2 water bottles

you can make it

5.625 cups = 4.5 water bottles

Then, all you have to do is divide both sides of this equation by 4.5, so

5.625 / 4.5 = 1 water bottle

1.25 cups = 1 water bottle

7 0

3 years ago

Simplify: 20 - (-2) to the second power ÷ ( 7 - 9) × 13

tresset_1 [31]

Answer:

The answer to the statement provided is 46.

4 0

3 years ago

Don't guess!

nekit [7.7K]

Answer:

Step-by-step explanation:

First off, we know that Trinity is allowed to watch 420 minutes of television a week, which means that her watch time can be=420, cannot be >than 420 minutes, and can be <than 420 minutes. (in other words, her watch time can be equal to or less than 420 minutes) Because of this, we know that our answer choice must have 420≥. So H and J are out of the question.

Also, we know that each show that she watches is 30 minutes long. Because of this, the next part of our answer is going to be 30s. (I will explain this in the comment section of my post because it's getting a little long)

Therefore, the answer is G: 30s≤420

8 0

3 years ago

Read 2 more answers

8x-3y times 5x5y over 10x6y2

erica [24]

Answer:

8x-5/4

Step-by-step explanation:

Cancel 5y x 5y/10 x 6y²=5/12y

8x-3y 5/12y

apply fraction rule a× b/c = a×b

____

8x-3y⁵/12y

8x-3y×5/12y

8x-3×5/12y

8x-3×5/12

=8x-5/4

6 0

2 years ago

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mea

QveST [7]

A.
$\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)$

Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.

$\mathbb P(Z>-2)=\mathbb P(|Z|2)=0.95+0.025=0.975$

b.
$\mathbb P(1100$

About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.

c. You're looking for $k$ such that

$\mathbb P(X>k)=0.10$

Since

$\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10$

occurs for $k^*\approx1.2816$ , it follows that

$\dfrac{k-1200}{100}=1.2816\implies k\approx1328$

So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.

3 0

4 years ago