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3 years ago

What is the answer!???

Mathematics

1 answer:

mr Goodwill [35]3 years ago

3 0

I believe A is the answer!

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How do I solve this? x+1/x-2 > 0

Eduardwww [97]

Answer:

x ∈ (-oo ; - 1) ∪ (+ 2 ; +oo)

Step-by-step explanation:

$\frac{x+1}{x-2}>0\\ \\x+1>0=>x>-1\\\\x-2>0=>x>2$

x ∈ (+ 2 ; +oo)

$x+1x$

x ∈ (-oo ; - 1)

x ∈ (-oo ; - 1) ∪ (+ 2 ; +oo)

8 0

3 years ago

Mishka is on a long road trip, and she averages 75 mph for 2 hours while she's driving on the highway. While she's driving on si

slega [8]

If she goes 75 mph for 2 hours, that's 150 miles altogether. 45 mph for 1 hour is 45 miles, so 150 + 45 is 195 miles.

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3 years ago

the krazy kite company is planning to produce kites.the company has rented space for its manufacturing operation at $6,000 per m

Leviafan [203]

37k (cost of the materials for each kite) plus 28k (the labor for each kite) plus 6,000=x (the total cost). With this equation you can plug in any number of kites so...

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65k+6,000=x Plug in k...

65(170)+6,000=x Multiply...

11,050+6,000=x Add...

x=17,050

4 0

3 years ago

How many times dose 28 go into 6

Ivan

6 goes 4 times into 28 because 6 x 4 = 24

3 0

3 years ago

Read 2 more answers

A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no mor

Juliette [100K]

Answer:

The 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

Let X = number of boards that fall outside the most rigid level of industry performance specifications.

In a random sample of 300 boards the number of defective boards was 12.

Compute the sample proportion of defective boards as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - α)% confidence interval for population proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of z for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

3 0

3 years ago