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2 years ago

Solve the inequality x + 1 > 5. Then graph the solutions.

1 answer:

4 0

Solution to Inequality #1:

6 - x > 15
-x > 15 - 6
-x > 9
-x/-1 < 9/-1 (divide by -1; reverse sign of inequality)
x < -9

Solution to Inequality #2:

8x < 1
x < 1/8

OR

x - 9 > -5
x > -5 + 9
x > 4

Solution to Inequality #3:

|x + 3| >= 4

Positive Case:

x + 3 >= 4
x >= 4 - 3
x >= 1

Negative Case:

-1(x + 3) >= 4
-x - 3 > = 4
-x >= 4 + 3
-x >= 7
-x/-1 <= 7/-1 (divide by -1; reverse sign of inequality)
x <= -7

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The percentage of water in an unknown hydrate was determined by heating the sample and driving the water off the sample. Two ind

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Answer:

It should be reported as 20.648%

Step-by-step explanation:

Since we have 2 different observed values hence we shall use an average of the 2 values to report the result

Thus value is $\frac{19.564+21.731}{2}=20.6475%$

As we can see that the least count of our observations is upto 3 decimal places hence we have to report a result upto only 3 decimal places thus we need to round off the fourth decimal place thus the digit shall be increased by 1 since we have to drop off 5 and the digit before 5 is 7 which is an odd number.

Thus the result shall be 20.648%

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2 years ago

What is the answer to y=3/4x+2 ???

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The answer is X=8/3

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2 years ago

Express p in terms of q if

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$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

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2 years ago

In preparation for an earnings report, a large retailer wants to estimate p= the proportion of annual sales

mr Goodwill [35]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

The sample size and the estimate are given by:

$n = 161, \pi = \frac{37}{161} = 0.2298$

Hence:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 - 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.1648$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 + 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.2948$

The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

More can be learned about the z-distribution at brainly.com/question/25890103

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