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2 years ago

Solve the inequality 1/4m≤−17 .

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

8 0

Answer:

$M\leq -68$

Step-by-step explanation:

This inequality is quite simple to solve.

Just multiply both sides by four and you have your answer:

$4*(1/4m)\leq4*-17$

$m\leq -68$

Hope this helps!

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A. 3 B. 3.1 C. 3.8 D. 11.6

pishuonlain [190]

Mean = (2+4+6+5+2)/5 = 19/5 = 3.8

answer

C. 3.8

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3 years ago

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ivolga24 [154]

Answer:

In the first box, put 24y

In the second box, put 16

Step-by-step explanation:

3y x 8 = 25y

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Hope that helps! :)

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6 0

3 years ago

Read 2 more answers

Find the Missing Numerator and Denominator

n200080 [17]

Answer:

missing denominator: 5

missing numerator: 3

Step-by-step explanation:

You know the product of fractions is formed by multiplying numerators and multiplying denominators.

$\dfrac{3}{d}\times\dfrac{n}{4}=\dfrac{3n}{4d}=\dfrac{9}{20}$

This gives rise to two equations:

3n = 9 ⇒ n = 3

4d = 20 ⇒ d = 5

The missing denominator is 5; the missing numerator is 3.

$\dfrac{3}{5}\times\dfrac{3}{4}=\dfrac{9}{20}$

7 0

2 years ago

Solve the proportion.3/4=v/14

Brums [2.3K]

Answer:

10.5

Step-by-step explanation:

3 0

3 years ago

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that $n = 1067, \pi = \frac{74}{1067} = 0.069$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079$

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0

3 years ago