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3 years ago

I need HELP ASAP!! It's for a quiz

Mathematics

2 answers:

ss7ja [257]3 years ago

7 0

Answer:The answer is odd

Alexus [3.1K]3 years ago

5 0

Answer:

even

Step-by-step explanation:

even because they are all even number for f , Theres megatives and positives yes I know but the number are still even!

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PLEASE HELP SOON I ONLY NEED HELP WITH NUMBER 18 PICTURE IS INCLUDED

anyanavicka [17]

Critical thinking? We're in trouble. It's hard for me to tell what grade this is.

Stem and leaf is fine but let's just write out the data.

Rich: 35 37 41 42 43 44 45 48 50 55 n=10

Will: 42 43 44 44 46 47 47 48 49 50 n=10

Use measures of center and measures of variation ...

The three measures of center typically used are the mean, the median and the mode.

Rich's sum = 440 so Rich's mean = 440/10 = 44

Rich's middle values are 43 and 44 so Rich's median = 43.5

All Rich's scores are unique so he doesn't really have a mode.

Will's sum = 460, Will's mean = 460/10 = 46

Will's middle numbers are 46 and 47 so median 46.5

Will's data has two modes tied, 44 and 47.

OK, so far these say Rich is a better golfer; he has lower mean and median scores than Will. (Low score is better in golf.)

Let's look at measures of variation. I'm guessing that they don't want to you calculate standard deviation. So we'll do range and mean absolute deviation.

The range is max-min.

Rich's range is 55-35=20

Will's range is 50-42=8

Will is more consistent.

Rich's mean is 44, his scores are here; let's calculate absolute deviation:

Score 35 37 41 42 43 44 45 48 50 55

AbsDev 9 7 3 2 1 0 1 4 6 11

Rich's deviations add to 44 so his MAD is 4.4

Will's mean is 46

Score: 42 43 44 44 46 47 47 48 49 50

AbsDev 4 3 2 2 0 1 1 2 3 4

The sum is 22, Will's MAD is 2.2

Based on the measures of variation, Will is a more consistent golfer. He's on average worse than Rich, but he always manages to stay pretty close to his average while Rich is all over the place.

Still I'd go with Rich for the tournament.

7 0

3 years ago

The bumper car ride at the state fair has 2 red cars, 4 green cars, an for the ride and is assigned a the probability that both

ollegr [7]

Answer:

a) 3.6%

Step-by-step explanation:

The given question mixed up, below is the correct question:

The bumper car ride at the state fair has 2 red cars, 4 green cars, and 2 blue cars. Garth is first in line for the ride and is assigned a car at random. Patty is next in line and is randomly assigned a car. Find the probability that both events A and B occur. Express your answer as a percent. If necessary, round your answer to the nearest tenth.

Calculation:

Given that the state fair has 2 red cars, 4 green cars and 2 blue cars.

There are therefore 2+4+2 = 8 cars in total.

Probability that Events A occurs P(A) = $\frac{2}{8}$ = 4

Probability that Events B occurs P(B) = $\frac{1}{7}$

Probability that Events A and B occur P(A ∩ B) = $\frac{2}{8}$ × $\frac{1}{7}$ = $\frac{2}{56}$ = 0.0357 = 3.57% ≈ 3.6%

Therefore, the probability that both events A and B occur is 3.6%

5 0

4 years ago

A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 45 pounds each, and the small boxes

shutvik [7]

Answer:

50(115-y) + 25y = 4125 [from eq2]

5750 - 50y + 25y = 4125 [distribute]

5750 - 25y = 4125

-25y = -1625

y = 65 [divide both sides by (-25)]

There are 65 small boxes.

Put that value into either equation (now, which is easier?) to solve for x:

x = 115 - y

x = 115 - 65

x = 50

There are 50 large boxes.

- Mring0506

4 0

4 years ago

Congruent figures are figures where...(choose all that apply)

ElenaW [278]

3 is the answer because the a correct

6 0

3 years ago

WHat is derivative of x^x^x^x

kicyunya [14]

Answer:

The answer is 0

Step-by-step explanation:

Technically you didn't term x, so it would be a constant with respect to x

5 0

4 years ago