Someone i really need no one ever helps me please someone help. And if i dont finish i am gonna get in trouble

Jason wants to plant 66 carrot seeds and 44 onion bulbs. He would like to plant them so

dexar [7]

The greatest number of plants he can have in each row is 22

He would have 3 rows of carrots

He would have 2 rows of onions

<h3>Greatest Common Factor</h3>

From the question, we are to determine the greatest number of plants he can have in each row

From the given information,

Jason wants to plant 66 carrot seeds and 44 onion bulbs

The greatest number of plants he can have in each row is the greatest common factor of 66 and 44

Factors of 66: 1, 2, 3, 6, 11, 22, 33, and 66

Factors of 44: 1, 2, 4, 11, 22 and 44

The greatest common factor of 66 and 44 is 22

Thus,

The greatest number of plants he can have in each row is 22

He would have 66/22 rows of carrots

66/22 = 3

Thus,

He would have 3 rows of carrots

He would have 44/22 rows of onions

44/22 = 2

Hence, he would have 2 rows of onions

Learn more on Greatest Common Factors here: brainly.com/question/5535901

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3 0

1 year ago

Describe two ways you can use the table to write a proportion

ale4655 [162]

You can use a table like this x -1 0 1 2
y 0 2 4 6
and a T table

4 0

3 years ago

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Determine the solution to the following system of equations: 2x+4y=-12 x+2y=-2

jolli1 [7]

Answer:

undefined/no solution

Step-by-step explanation:

you use elimination to solve. So multiply 2x+4y=-12 by 1 and x+2y=-2 by 2. then you subtract 2x+4y=-12 by 2x+4y=-4 and get 0y=-8. Anything divided by 0 is undefined/no solution because you can't divide anything by 0.

4 0

3 years ago

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What is the volume of the box? Use unit cubes. Click the unit cube button below.

Sonbull [250]

Answer:

72ft^3

Step-by-step explanation:

if you put the 6 by 3 unit cubes thing you will fill in the box with 4 of them do 18 X 4 = 72

8 0

2 years ago

Read 2 more answers

Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of

Ket [755]

Answer:

A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.

B) P = 0.7616

C) P = 0.4441

Step-by-step explanation:

We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.

A) We take a sample of size n=50.

The mean of the sampling distribution is equal to the population mean:

$\mu_s=\mu=18.5$

The standard deviation of the sampling distribution is:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849$

B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.

We can calculate this with the z-scores:

$z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179$

The probability it then:

$P(|X_s-\mu_s|$

C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589$

$P(|X_s-\mu_s|$

5 0

3 years ago