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2 years ago

Is the value of the fraction 7−2y/6 greater than the value of the fraction 3y−7/12 ?

Mathematics

1 answer:

Sveta_85 [38]2 years ago

3 0

Complete Question:

Is the value of the fraction 7−2y/6 greater than the value of the fraction 3y−7/12 ? For what values?(Make sure to use an inequality)

Answer:

y < 3

Step-by-step explanation:

The given two fractions are:

$\frac{7-2y}{6}$ and $\frac{3y-7}{12}$

We have to tell for which range of values is the value of first fraction larger than the second fraction. This can be done by setting up an inequality as shown:

$\frac{7-2y}{6} >\frac{3y-7}{12}$

The range of y which will satisfy this inequality will result in first fraction of larger value as compared to the second fraction.

Multiplying both sides of inequality by 12, we get:

$12 \times \frac{7-2y}{6} > 12 \times \frac{3y-7}{12} \\\\ 2(7-2y)>3y-7\\\\ 14-4y>3y-7\\\\ 14+7>3y+4y\\\\ 21>7y\\\\ 3>y\\\\ y$

This means, for y lesser than 3, the value of first fraction is larger than the second one.

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Read 2 more answers

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a. The test statistic for this hypothesis would be z=1.71

b. Critical value at α = 0.10: z=1.282

c. Reject H0, the television network should not keep its current lineup.

d. H0: p ≤ 0.50; HA: p > 0.50

Step-by-step explanation:

We have to performa an hypothesis test on a proportion.

The claim is that the proportion of viewers that prefer the new show is bigger than 50% (meaning that most viewers prefer the new show).

The null hypothesis will state that both shows have the same proportion.

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$H_0: \pi\leq0.50\\\\H_a: \pi>0.5$

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Then, the z-statistic can be calculated as:

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The critical value for a right tailed test at a significance level of 0.10 is zc=1.282. This value can be looked up in a standard normal distribution table.

As the statistic is bigger than the critical value, it lies in the rejection region. The null hypothesis is rejected. There is evidence to support the claim that the proportion of viewers which support the new show is larger than 0.50.

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