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Arlecino [84]
3 years ago
9

It cost Jocelyn $5.60 to send 56 text messages. How much would it cost to send 198

Mathematics
2 answers:
victus00 [196]3 years ago
7 0

Answer:

$19.8

Step-by-step explanation:

To find the answer, find the  ratio.

Sloan [31]3 years ago
4 0

Answer: 1980

Step-by-step explanation:

Set yp a proportion

5.60dollars=56texts

----------------   ----------

198dollars=    x

Then you cross multiply the 198 with the 56

then you divide the 5.60

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Suppose a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%. A cell phon
leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = <u><em>proportion of households that have three cell phones NOT known.</em></u>

So, Null Hypothesis, H_0 : p = 30%      {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, H_A : p \neq 30%     {means that the actual percentage of households different from 30%}

The test statistics that would be used here <u>One-sample z-test for proportions;</u>

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

where, \hat p = sample proportion of households having three cell phones = \frac{43}{150} = 0.29

           n = sample of households = 150

So, <u><em>the test statistics</em></u>  =  \frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }  

                                     =  -0.27

The value of z test statistic is -0.27.

<u>Also, P-value of the test statistics is given by;</u>

              P-value = P(Z < -0.27) = 1 - P(Z \leq 0.27)

                            = 1 - 0.6064 = <u>0.3936</u>

<u>Now, at 1% significance level the z table gives critical value of -2.58 and 2.58 for two-tailed test.</u>

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <em><u>we fail to reject our null hypothesis</u></em>.

Therefore, we conclude that the actual percentage of households is equal to 30%.

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3 years ago
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