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2 years ago

The expression 4c gives the science test score that a student recives c correct problems. what score is for 18 correct problems

Mathematics

1 answer:

Shtirlitz [24]2 years ago

5 0

4c
4(18)
72 is the score for 18 correct problems

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How much

morpeh [17]

Answer:

She would need at least 11 buses leaving 17 spare seats.

Step-by-step explanation:

171+22+13= 206

206/20= 10.3

10.3 = 11

3 0

1 year ago

In the diagram below, P and Q are points on a circle with centre O.

Gnoma [55]

Answer:

Step-by-step explanation:

According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.

Also from the diagram, ∠OQP + ∠PQT = ∠OQT;

∠PQT = ∠OQT - ∠OQP

Given ∠OQP = 18° and ∠OQT = 90°

∠PQT = 90°-18°

∠PQT = 72°

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3 years ago

A woman earns 2600 per month and budgets 468 per month for food what percent of her monthly income is spent on food

finlep [7]

divide food money by income:

468 /2600 = 0.18 = 18%

3 0

3 years ago

Angie’s Bake Shop makes birthday chocolate chip cookies that cost $2 each. Angie expects that 13% of the cookies will crack and

Stella [2.4K]

I think it is $287....................

6 0

2 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

2 years ago