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3 years ago

How do I solve Proportions?

Mathematics

2 answers:

Bumek [7]3 years ago

6 0

Proportions is an what to 1 as in 1 seed to every pumpkin

frosja888 [35]3 years ago

6 0

Cross multiply and then if they are the same they are proportional if not they are not

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The revenue, in thousands of dollars, of a new flavor of liquor is given below as function of the number of gallons (in hundreds

Naddik [55]

Answer:

$625,000

Step-by-step explanation:

The maximum revenue will be found where the derivative of the revenue function is zero.

r'(x) = 50 -2x = 0

50 = 2x

x = 25

When 2500 gallons are sold, the revenue is ...

r(25) = 50(25) -25(25) = (50 -25)(25) = 25(25) = 625 . . . thousands

The maximum revenue is $625,000.

8 0

2 years ago

The correlation analysis assumes that the measurements have a bivariate normal distribution in the population. Select all of the

hoa [83]

Answer:

Bruh i got correlation hw that i posted on here too and nobody helped lol.

Step-by-step explanation:

8 0

3 years ago

there is a line that includes the point (2,2) and has a slope of 3. What is its equation in slope intercept form?

Nana76 [90]

Answer:

y = 3x - 4

Step-by-step explanation:

y = 3x + b

2 = 3(2) + b

2 = 6 + b

-4 = b

y = 3x - 4

5 0

2 years ago

Whoever is the 1st one to answer i u mark u as brainliest

Vaselesa [24]

Answer:

the answer is -1... since you did 2 negative numbers

GIVE ME BRAINLIEST PLZZZZ

Step-by-step explanation:

8 0

3 years ago

A sociologist wishes to estimate the percentage of the United States population living in poverty. What size sample should be ob

andriy [413]

Answer:

a) We need a sample size of at least 705.

b) We need a sample size of at least 1692.

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 $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

a) If she uses 1999 estimate of 11.8% obtained from the Current Population Survey.

We need a sample of size at least n

n is found when $M = 0.02, \pi = 0.118$ .

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

$0.02 = 1.645\sqrt{\frac{0.118*0.882}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.118*0.882}$

$\sqrt{n} = \frac{1.645\sqrt{0.118*0.882}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.118*0.882}}{0.02})^{2}$

$n = 704.08$

Rounding up

We need a sample size of at least 705.

b) She does not use any estimate.

Same thing as above, we just use $\pi = 0.5$ when do not use any estimate.

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

$0.02 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.5*0.5}$

$\sqrt{n} = \frac{1.645\sqrt{0.5*0.5}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.5*0.5}}{0.02})^{2}$

$n = 1691.2$

Rounding up

We need a sample size of at least 1692.

4 0

3 years ago