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2 years ago

I need help with this

Mathematics

1 answer:

choli [55]2 years ago

6 0

Answer:

Step-by-step explanation:

Take what's given and write an equation. If you know that the distance to the mid point is half of the entire length, you can say

2(x+6) = 9x -2

Distribute to get 2x + 12 = 9x -2

Solve 12 +2 = 9x -2

7x = 14. So x=2

Substitute 2 for x in the original equation.

2(2+6) = 9(2) - 2

16 = 16

So the length of UW is 16

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<h3>How to calculate the percentage?</h3>

From the information given, the total amount spent will be:

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1 year ago

What is the volume of the rectangular prism?<br> 1/3ft

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2 years ago

If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (

AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

$P=\frac{3}{36}$

$P=\frac{1}{12}$

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

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2 years ago

SUSTITUTION 3X-6Y=2<br> 5X+4Y=1

Montano1993 [528]

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(-1/3, -1/6)

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