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kkurt [141]
3 years ago
15

7.RP.A.1 3. If 1 1/2 cups of something fills 2/3 of a container, how many cups fill the container? ​

Mathematics
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

Hi there!

Your answer is:

2 & 1/4 <u>OR</u> 2.25 cups fills the container!

Step-by-step explanation:

If 1 & 1/2 (aka 3/2) fills 2/3 of the container, then:

we know that <em>half</em> of 3/2s fills 1/3 of the container

To find half of 3/2, we multiply 3/2 by the <em><u>inverse</u></em> of 2, which is 1/2.

3/2÷2

3/2×1/2 = 3/4 ths

Now that we know 3/4ths fills 1/3 of the container, we add that to 1& 1/2.

3/4+ 1&1/2 = 2 & 1/4

Check your work!

3/4 × 3 should equal 2&1/4 if we are correct

3/4 × 3/1 = 9/4.

Simplify!

9/4 = 2&1/4

That tells us that <u>we</u><u> </u><u>are</u><u> </u><u>correct</u><u>!</u>

Hope this helps!

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NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400
Ne4ueva [31]

Answer:

The test statistics is  t  =  -1.727

Step-by-step explanation:

From the question we are told that

The data given is  

   330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

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    The  sample  size is  n =  16

  The  null hypothesis is  \mu \le  14400

    The  alternative hypothesis is  H_a  :  \mu > 14400

The sample mean is mathematically evaluated as

  \= x  =  \frac{\sum x_i}{n}

So

   \= x  =  \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}

=>  \= x = 10799.9

The  standard deviation is mathematically represented as

      \sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}

So

\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}}  \ ..

   ..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2  }{16}} \ ...

  \ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}

=>  \sigma  =  8340

  Generally the test statistic is mathematically represented as

  t =  \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }

t  =  -1.727

From the z-table  the p-value is  

     p-value  = P(Z > t) =  P(Z >  -1.727) =  0.95792

 From the values obtained we see that

        p-value  >  \alpha  so we fail to reject the null hypothesis

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