Pythorgorean Theorem

Mathematics

1 answer:

mote1985 [20]3 years ago

3 0

Are you sure it is Pythorgoren theorem because I was search for the answer but I could not find the answer this all it shows me

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Maslowich

The graph that represents viable values for y = 2x is Option A.

<h3>What is a Straight Line Function ?</h3>

A straight line function is given by y = mx +c , where m is the slope and c is the y intercept.

The given equation is y = 2x

here m = 2

x is the number of pounds of rice scooped and purchased from a bulk bin

y is the total cost of the rice

as both the data cannot be negative , Option C , D is out of choice

The Option 1 represents a straight line and it starts at the origin which is satisfied by y = 2x as y = 0 , at x = 0

ends at point (2.5, 5) giving a slope of m =2 ,

Therefore , The graph that represents viable values for y = 2x is Option A.

To know more about Straight Line equation

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1 year ago

You deposit $500 in an account that earns 4% simple annual interest entrance and each year is added to the principal to create a

ikadub [295]

Answer:

first year $20 second year $20.80 Third year $21.63

Step-by-step explanation: This is for an extra money per year so after the first year that is how much extra money he gets every year.

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2 years ago

A quadrilateral has three angles which measure 23.5 degrees, 99 degree, and 100.4 degrees. What is the measure of the fourth ang

Schach [20]

I think it would be 137.1, due to the total being 360. Just subtract.

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3 years ago

Read 2 more answers

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).