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2 years ago

True or False.The following relation represents a function.

Mathematics

1 answer:

enot [183]2 years ago

4 0

True because each point has it's own individual column

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Use the figure to find the measure of each arc for #1-4.

Bond [772]

Answer:

Step-by-step explanation:

6 0

1 year ago

I need help with this problem. asap.

Amanda [17]

Answer:

dang I don't know! All I know is that it is more than 1-2 inches... I think.

6 0

2 years ago

Read 2 more answers

5b= -G+ 1/5 UA Solve for a:

enot [183]

Answer:

switch sides

-g+2/5UA=5b

add g to both sides

-g+1/5UA+g=5b+g

simplify

1/5UA=5b+g

multiply both side by 5

5×1/5UA=5×5b+g

simplify

AU=25b+5g

divide both sides by U

AU/U=25b/u+5g/u =U/U=0

simplify

A=25b+5g/U

8 0

3 years ago

I will mark brainliest and give 30 points

zhuklara [117]

Its B, The answer for your question is B

4 0

2 years ago

Read 2 more answers

assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the

drek231 [11]

Answer:

65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value $x_{0}$ such that $P(X < x_{0}) = 0.76$ , but, $0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5)$ , i.e., $(x_{0}-63.6)/2.5$ is the 76th percentile of the standard normal distribution. So, $(x_{0}-63.6)/2.5 = 0.7063$ , $x_{0} =63.6+(2.5)(0.7063) = 65.3658$ . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.

7 0

3 years ago

True or False.The following relation represents a function. ​

True or False.The following relation represents a function.