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slega [8]
3 years ago
15

What do I need to multiply by to decrease by 42%?

Mathematics
2 answers:
Arisa [49]3 years ago
7 0
Hi there!~

So first off we need to use a variable to solve the given percentage, lets use "y" (This can also be used as any other letter, most commonly can be known as "x")

The formula is "\frac{100-y}{100}"

And because we are increasing, we would use that ^

So for us we get 0.58 as an answer. 
Leona [35]3 years ago
4 0
32.88.36 72 thats your awnser
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What is the equation of this line?
Svetlanka [38]

Answer:

A y=-2x-2 I think so sorry if wrong

8 0
3 years ago
Are same side interior angles congruent?
Fed [463]
Same side interior angles are not always congruent
6 0
4 years ago
Read 2 more answers
Solve the system 2x+4y=16 5x +3y=19 then find the value of y/x
AURORKA [14]

Answer:

I would change 2x+4y=24 into x=12–2y

To do that, divide both sides by 2 and then subtract 2y on each side.

After that, substitute for x. 3x+2y=19 would become 3(12–2y)+2y=19.

Then solve.

3(12–2y)+2y=19

36–6y+2y=19

-4y+36=19

-4y=-17

y=4.25

Then, substitute y in either equation.

Either this:

3x+2y=19

3x+2(4.25)=19

3x+8.5=19

3x=10.5

x=3.5

Or:

2x+4y=24

2x+4(4.25)=24

2x+17=24

2x=7

x=3.5

Or you could solve it in the equation you created in the beginning:

x=12–2y

x=12–2(4.25)

x=12–8.5

x=3.5

The coordinates where the lines intercept are (3.5, 4.25).

Sorry for the long answer!

4 0
3 years ago
An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.
andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

6 0
3 years ago
Help me please asap
weeeeeb [17]
C is the answer to you question
3 0
3 years ago
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