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2 years ago

Please please please help me with this answer

Mathematics

1 answer:

aleksandrvk [35]2 years ago

8 0

57-29=28 so there is 28 green peppers

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Pleas answer part A!!!!!!

FrozenT [24]

Answer:

y=120x

Step-by-step explanation:

You can just divide all of the points (y/x) and they all equal to 120. And when you add it to the equation its y=120x

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2 years ago

the dot plots how the salaries for the employees at two small companies before a new company head is hired at each company. Sele

leonid [27]

Answer:

I need a picture to solve this.

Step-by-step explanation:

4 0

2 years ago

It's question B as I've done A please help

Hatshy [7]

Its easy.use cos rule.

4 0

2 years ago

the city of brice is planning to build a skate park. An architect designed the area shown at the right. in the plan, the perimet

Amiraneli [1.4K]

The actual skate park's perimeter is 130 inches.

Explanation:

Step 1; Assume the initial garden has a width of y inches. It is given that the length is 25 inches. The perimeter of any given rectangle is twice the sum of the length and the width of the same rectangle. The initial perimeter is given as 80 inches.

Perimeter = 2 × (length of the rectange + width of the rectangle).

80 = 2 × (25 + y), 40 = 25 + y, y = 40 - 25 = 15

So the initial park has a width of 15 inches.

Step 2; Now we calculate the actual skate park's perimeter. The length is given as 50 inches and the width was found to be 15 inches.

Perimeter = 2 × (length of the rectange + width of the rectangle).

Perimeter = 2 × (50 + 15) = 2 × 65 = 130 inches.

5 0

2 years ago

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden

hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is $\hat{p}=26/42=0.6190$ . The estimated standard deviation is given by $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749$ . Because we have a large sample, the 90% confidence interval for p is given by $0.6190\pm z_{0.05}0.0749$ where $z_{0.05}=1.6448$ is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is $0.6190\pm (1.6448)(0.0749)$ , i.e., (0.4958, 0.7422).

5 0

2 years ago