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djyliett [7]
3 years ago
14

In 45 years, Gabriela will be 4 times as old as she is right now. How old is she right now?

Mathematics
1 answer:
zhenek [66]3 years ago
8 0

Answer:

15 yrs. old  

Step-by-step explanation:

x+45=4x     45=3x      15=x  

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Helppppppppppppejejehehehrhe
tino4ka555 [31]

Answer:

4/5 and the second one is 3

Step-by-step explanation:

the fact that you dont, know this is sad.

8 0
3 years ago
What is -x-3y=-3 and 2x+3y=5 using elimination method?
xxTIMURxx [149]

Answer:(2,1/3)


Step-by-step explanation:

-x-3y=-3

2x+3y=5


_________First solve for x since y is 0. -3y+3y=0

1X = 2

1x/1= 2/1

X= 2

Now substitute your x answer into any equation to find y I used second equation

2x+3y= 5

2(2)+3y=5

4+3y=5

3y=5-4

3y=1

3y/3=1/3

Y=1/3



6 0
4 years ago
M
irakobra [83]

Step-by-step explanation:

now even though I cannot find most of the answers for you I can tell you how to solve them, all you need to do is just input the equations that equal the letter pairs then sold them as if it was a find the x equation but in this case we're solving for L and not x

4 0
3 years ago
A sample of 900900 computer chips revealed that 66f% of the chips fail in the first 10001000 hours of their use. The company's p
Alik [6]

Answer:

No, there is not enough evidence at the 0.02 level to support the manager's claim.

Step-by-step explanation:

We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.

And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;

Null Hypothesis, H_0 : p = 0.68 {means that the actual percentage that fail is same as the stated percentage}

Alternate Hypothesis, H_1 : p  0.68 {means that the actual percentage that fail is different from the stated percentage}

The test statistics we will use here is;

     T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } } ~ N(0,1)

where, p = actual percentage of chip fail = 0.68

            \hat p = percentage of chip failed in a sample of 900 chips = 0.66

           n = sample size = 900

So, Test statistics = \frac{0.66 -0.68}{\sqrt{\frac{0.66(1- 0.66)}{900} } }

                             = -1.267

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.

Therefore, we conclude that the actual percentage that fail is same as the stated percentage and the manager's claim is not supported.

5 0
4 years ago
35x68<br> What is the answer
timurjin [86]

Answer:

2,380

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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