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1 year ago

The ages of five friends are 25, 36, 33, 36 and 30 how many of friends are older than the average (arithmetic mean) age?

Mathematics

1 answer:

Murrr4er [49]1 year ago

4 0

Answer:

3 friends are older than the average.

Step-by-step explanation:

First find the mean. The mean can be found adding all the numbers in the set together and dividing by the amount of numbers in the giving set:

(25 + 36 + 33 + 36 + 30)/5 = (160)/5 = 32

The average age is 32. There are 3 friends that are older than the average (33 , 36 , 36, respectfully).

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A triangle has a sides (x) feet, (x+3) feet, and (x+5) feet. The perimeter is 99.5 feet What is the measure for the largest side

ASHA 777 [7]

Answer:

35.5 feet

Step-by-step explanation:

x + x + 3 + x + 5 = 99.5

3x + 8 = 99.5

3x = 91.5

x = 30.5

Largest side = (x+5)

30.5 + 5 = 35.5 feet

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2 years ago

A rabbit can run 35 miles per hour.A fox can run 21 miles in half an hour.Which animal is faster and how much

Pachacha [2.7K]

Answer:

Step-by-step explanation:

The fox is faster. Because he is a fox.

Also because he can (assuming he doesn't fatigue or get caught by hounds in the second half of the hour) run 42 miles in 1 hour.

The rabbit runs 35 miler per hour.

Velocity = Distance/time

Fox velocity = 21 miles/0.5 hour

Dividing by 0.5 is the same as multiplying by 2, so...

Fox velocity = 42 miles/hour

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2 years ago

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AlladinOne [14]

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3 years ago

Read 2 more answers

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: