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3 years ago

What are the solutions of the compound inequality 2d + 3 ≤ –11 or 3d – 9 > 15?

Mathematics

1 answer:

jolli1 [7]3 years ago

7 0

Answer:

A. d ≤ –7 or d > 8.

Step-by-step explanation:

Given : 2d + 3 ≤ –11 or 3d – 9 > 15.

To find : What are the solutions of the compound inequality .

Solution : We have given 2d + 3 ≤ –11 or 3d – 9 > 15.

For 2d + 3 ≤ –11

On subtracting both sides by 3

2d ≤ –11 - 3 .

2d ≤ –14.

On dividing both sides by 2 .

d ≤ –7.

For 3d – 9 > 15.

On adding both sides by 9.

3d > 15 + 9 .

3d > 24 .

On dividing both sides by 3 .

d > 8 .

So, A. d ≤ –7 or d > 8.

Therefore, A. d ≤ –7 or d > 8.

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Read 2 more answers

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 = proportion of households that have three cell phones NOT known.

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 One-sample z-test for proportions;

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, the test statistics = $\frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

= -0.27

The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

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

= 1 - 0.6064 = 0.3936

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

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 we fail to reject our null hypothesis.

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

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3 years ago