What is the radius and diameter of the following circle?

Suppose j varies jointly with g and v, j=1 when g=6 and v=3. Find j when g=8 and v=11

Lady bird [3.3K]

j=4.88 when g=8 and v=11

Further explanation:

When the increase/decrease in one quantity cause increase/decrease in other quantity, it is called direct variation.

Variation is always accompanied by a variation constant.

<u>Given</u>

g and v vary directly with j

IT can be written as:

j∝gv

Putting the variation constant k

j = kgv

Putting g = 6 and v=3

$j = kgv\\1 = k * 6 * 3\\1 = 18k\\k = \frac{1}{18}$

So the value of k is 1/18 which makes the equation

$j = \frac{1}{18}gv\\ Putting\ g=8\ and\ v=11\\j = \frac{1}{18} * 8 * 11\\j = \frac{88}{18}\\ j =4.88$

So, j=4.88 when g=8 and v=11

Keywords: Variation, Direct Variation

Learn more about variation at:

#LearnwithBrainly

8 0

3 years ago

a decimal number to say 23.7 and it was rounded give a decimal number that is less than 22.37 and another that is greater than 2

chubhunter [2.5K]

23.7 IS GOING TO BE RIGHT BECAUSE 23.73 ROUNDS TO 23.7

6 0

3 years ago

Read 2 more answers

Calculate the area of the circle , round your answer to the nearest hundredth

lesantik [10]

The area is 25π

Hope this helps

3 0

4 years ago

Thirty-seven percent of all Americans drink bottled water more than once a week (Natural resources Defense Council, December 4,

lesantik [10]

Answer:

a) The sampling distribution for n=540 has a mean sample proportion of p=0.37 and a standard deviation of σs=0.0208.

b) probability = 0.99998

d) probability = 0.99874

e) You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

Step-by-step explanation:

a) We have a known population proportion π=0.37 and we have to describe the sampling distribution when the sample size is n=540.

The mean sample proportion is expected to be the same as the population proportion:

$\bar p = \pi = 0.37$

The standard deviation of the sampling will be the population standard deviation divided by the square root of the sample size:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{540}}=\sqrt{0.000431667}=0.0208$

Then, we can say that the sampling distribution will have a p=0.37 and a standard deviation σs=0.0208.

b) We have to calculate the probability that the sample proportion will be within 0.09 of the population proportion.

We can calculate the z-value as:

$z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0208}=4.3269\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0208}=-4.3269$

As the distribution is symmetrical, we can calculate the probabilty that he sample proportion will be within 0.09 of the population proportion as:

$P(|p-\bar p|$

probability = 0.99998

d. Now the sample is smaller (n=300), so the standard deviation of the samping distribution:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{300}}=\sqrt{0.000777}=0.0279$

We have to recalculate the z-scores:

$z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0279}=3.2258\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0279}=-3.2258$

And the probability is:

$P(|p-\bar p|$

probability = 0.99874

e. The increase in sample size is 80%

$\Delta n\%=\dfrac{n_1}{n_2}-1=\dfrac{540}{300}-1=1.8-1=0.8=80\%$

and the increase in probability is 0.12%

$\Delta P\%=\dfrac{P_1}{P_2}-1=\dfrac{0.99998}{0.99874}-1=1.0012-1=0.0012=0.12\%$

You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

7 0

3 years ago

a can of coffee weighs 13 ounces. what is the smallest number of cans that must be used to package 1,000 kilograms of coffee

Anna11 [10]

The point is that when you convert, you may get a fractional number, so you need to round up to the next can.

3 0

4 years ago