MATHEMATICS HIGH SCHOOL

A grocery store manager uses 1/2 crate of apples for every 3/4 crate if oranges in a fruit display. How many crates of oranges will she need if she uses 2 1/2 crates of apples?

Answers

Answer 1
Answer:

Answer:

15/4

Step-by-step explanation:

This is a proportion question

It says 1/2 crates of apple are used for every 3/4 crates of oranges

For 5/2 crates of apple, x crates of oranges will be needed.

To get x, we simply multiply 1/2 by x and equate it to 3/4 multiplied by 5/2

This means: 0.5x = 0.75 × 2.5

x = ( 0.75 × 2.5 ) ÷ 0.5 = 3.75

Hence for 5/2 crates for apple, she will need 15/4 crates of oranges.
Answer 2
Answer:

Answer:

Step-by-step explanation:

You would first make a proportion.

.5/.75 = 2.5/x

So, .5x = 1.875

x = 15/4

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Is this correct? I have tried and want it checking

Answers

Beautifully done !  Both questions.  Really.
You do know what you're doing.

The city park shown at the right has an area of 6889 m². The parks length along the front Street is 83 m. Find the length of the park along second Avenue.

Answers

length times width = area
83 m times w =6889
divide both sides by 83
and the second length w is 83m

Which type of lines pass through points (2, 5), (8, 7) and (-3, 1), (-2, -2) on a grid?PERP
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Answers

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Get this question correct image attached

Answers

Answer: x ≤ 5 or x > 7

Step-by-step explanation:

    First, we will write an equation for the left side. It's a closed dot going to the left (less than) of 5.

         x ≤ 5

    Next, we will write the right side. It's an open dot going to the right (greater than) of 7.

         x > 7

    Lastly, we will combine these to create our compound inequality.

         x ≤ 5 or x > 7

Answer:

-infinity less than or equal to 5 and 7 less than infinity

Step-by-step explanation:

On the distant planet Cowabunga, the weights of cows have a normal distribution with a mean of 483 pounds and a standard deviation of 70 pounds. The cow transport truck holds 5 cows and can hold a maximum weight of 2840. If 5 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 2840

Answers

Answer: 0.0033

Step-by-step explanation:

Let x be a random variable that denotes the weights of cows.

Given: \mu = 483,\ \sigma=70

maximum weight can be hold= 2840 pounds.

Mean weight = (2840)/(5) = 568 pounds

The probability their total weight will be over the maximum allowed of 2840

= P(X>2840)

P((x-\mu)/((\sigma)/(√(n)))>(568-483)/((70)/(√(5))))\n\n=P(z>2.715)\n\n=1-P(z<2.715)\n\n=1-0.9967=0.0033

Hence, the required probability = 0.0033
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