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3 years ago

The domain of the function is given. Find the range.

Mathematics

1 answer:

Flura [38]3 years ago

8 0

Answer:

your third answer

Step-by-step explanation:

its easy just plug in each domain into your function and the result will be the range

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I don’t know how to do this problem and need help getting answer?

3241004551 [841]

Hi!!! So to do this problem, you need to find the mean values for each sample. I will walk you through finding the mean of sample 1: first you want to add all of the values for sample 1, which are 4,5,2,4 and 3. Once you add those values you get 18. Then you must divide that 18 by the number of terms you added. The numbers you added were 4,5,2,4 and 3 like I said earlier which is 5 numbers. You divide 18 by 5 to get your mean, which is 3.6

Answer: (B) Sample 2

sample 1 mean = 3.6

sample 2 mean = 4.2

sample 3 mean = 3.8

sample 4 mean = 4

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3 years ago

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Molodets [167]

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3 years ago

9. What is the distance along the x axis from order pairs -8, 6 and 4,6

Blababa [14]

Answer:

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Step-by-step explanation:

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5 0

3 years ago

A construction worker needs to put a rectangular window in the side of a building. He knows from measuring that the top and bott

Sliva [168]

Answer: 5 feet

Step-by-step explanation: A rectangle is a quadrilateral with 4 sides. The opposite sides are parallel and have the same length.

Top and bottom of the window are opposite, so they have the same measure of 3 feet.

One length of the window is 5 feet. Since the last side is opposite to the length of the window, it has the same length, i.e., its value is 5 feet.

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3 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago