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3 years ago

HELP HELP HELP HELP ME PLEASE Answer 7 and 8

Mathematics

1 answer:

Svetach [21]3 years ago

8 0

I DONT KNOW FHUEITHNGIFJTFRJNRGV

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Hey! I'v been struggling with this question lately. Can someone just lend me a hand?

Dominik [7]

So, this problem is asking us to find the value of our variable u.

How can you do this? By isolating the variable, or having only it on one side of the equals sign.

Here's the step by step approach. It's just simple algebra and reverse order of operation (which is what you should do whenever you are solving for a variable).

$\frac{u+88}{11} = 17$

Multiply both sides by 11 to get rid of the 11 denominator.
$u + 88 = 187$

Then, subtract both sides by 88 to get u by itself.
$u = 99$

Therefore, u is equal to 99.

6 0

3 years ago

Read 2 more answers

Ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm help plz? :)

Daniel [21]

Answer:

1. factor

2. prime number

3. composite number

4. GCF

5. LCM

7 0

3 years ago

In politics, marketing, etc. We often want to estimate a percentage or proportion p. One calculation in statistical polling is t

inessss [21]

Answer: |p-72% |≤ 4%

Step-by-step explanation:

Let p be the population proportion.

The absolute inequality about p using an absolute value inequality.:

$|p-\hat{p}| \leq E$ , where E = margin of error, $\hat{p}$ = sample proportion

Given: A poll result of 72% with a margin of error of 4% indicates that p is most likely to be between 68% and 76% .

|p-72% |≤ 4%

⇒ 72% - 4% ≤ p ≤ 72% +4%

⇒ 68% ≤ p ≤ 76%.

i.e. p is most likely to be between 68% and 76% (.

6 0

2 years ago

how many gallons of a 50 antifreeze solution must be mixed with 90 gallons of 20% antifreeze to get a mixture that is 40% antifr

Tamiku [17]

Try this:
1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.
2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18).
3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of <u>pure</u> weight.
4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.
Answer: 108 gal.

3 0

3 years ago

−3 = −6 + 9<br> 2 + 4 = 12 can anybody help me with this ?

Ahat [919]

Answer:behave e is

Step-by-step explanation:

5 0

2 years ago