Determine whether the relation represents a function. Explain your reasoning. Your answer AY

1 answer:

6 0

A relation is a function if each element of the domain is paired with exactly one element of the range. ... If given a table, or a set of ordered pairs, you can look to see if any value of the domain has more than one corresponding value in the range.

tbh google

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Base your answer on Betty is thinking of two consecutive integers whose sum is 41.

mel-nik [20]

Answer:

let the numbers be, x and x+1 ( as it is consecutive number the next number will have 1 more than the number )

sum = 41

so, x + ( x + 1 ) = 41

x + x + 1 = 41

2x + 1 = 41

2x = 41 -1 = 40

2x = 40

x = 40/2 = 20

x = 20

x + 1 = 20 + 1 = 21

so the consecutive numbers whose sum is 40 are, 20 and 21

hope this answer helps you.....

please mark as brainliest... thank you!

6 0

2 years ago

Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, use the given sample sizes and numb

11Alexandr11 [23.1K]

Since the data is insufficient. Let us denote that n1 = 100, n2 = 100X1 = 38. X2 = 40
So looking for the p value:
p = (38+ 40) / (100 + 100) = 0.39
z = (38/100 - 40/100)/√ (0.39*(1-0.39)*(1/100 + 1/100))= -0.2899
P-value = P (|z| > 0.2899) = 0.7718

7 0

2 years ago

The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

Serhud [2]

Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

$C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi } = \sqrt{\frac{a^{2} + b^{2} }{2} }$

Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$