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3 years ago

15

A. 30 B.45 C. 90 D. 60

2 answers:

Lorico [155]3 years ago

7 0

Answer:

90

Step-by-step explanation:

BabaBlast [244]3 years ago

6 0

Answer:

<u>C. 90</u>

Step-by-step explanation:

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Find the value of x. Give reasons to justify your solution.

Alja [10]

Answer:

Answer attached

Step-by-step explanation:

steps and answer in attachment

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3 years ago

In 1983, a can of house paint cost $16.35. In 2000, a can of house paint cost $28.94, and in 2005, a can of house paint cost $32

Readme [11.4K]

The Consumer Price Index is calculated through the equation,
CPI = (current period price/base period price) x 100
The base period is 1983. Solving the CPIs for 2000 and 2005
(2000)
CPI = ($28.94/$16.35) x 100 = 177
(2005)
CPI = ($32.54/$16.35) x 100 = 199
The difference between the CPIs of 2005 and 2000 is 199 - 177 = 22.

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3 years ago

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HELP ASAP<br> G = a(4 – d)

NemiM [27]

Answer:

g/-d+4

Step-by-step explanation:

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3 years ago

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2x to the 3 + 6 if x = -3

alexdok [17]

Answer:

To my understanding the question is asking for this:

2x^(3+6) if x = -3

= 2(-3)^(3+6)

= 2(-3)^9

= -39366

Hope this helps!

If this is not the equation the question is asking for please rewrite your question.

6 0

3 years ago

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c

steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for difference between population proportions is:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23$

The critical value of <em>z</em> for 90% confidence interval is:

$z_{0.10/2}=z_{0.05}=1.645$

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)$

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0

3 years ago