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Over [174]
3 years ago
6

A football team was training for four hours. During the first hour, they practiced for 5/8 of an hour. During the second hour, t

hey practiced for 2/3 of an hour. During the last two hours, they first practiced for 3/5 of an hour, took a 1/2 hour break and then practiced the rest of the time. How much time did they spend practicing in total?
Mathematics
1 answer:
Leviafan [203]3 years ago
7 0

Answer: 2 19/24 hours was spent in practising.

Step-by-step explanation:

During the first hour, they practiced for 5/8 of an hour. During the second hour, they practiced for 2/3 of an hour. This means that the total time for which they practiced in the first 2 hours would be

5/8 + 2/3 = 31/24 hours

During the last two hours, they first practiced for 3/5 of an hour, took a 1/2 hour break and then practiced the rest of the time. This means that the rest of the time for which they practiced is

2 - (3/5 + 1/2) = 2 - 11/10 = 9/10

Therefore, the time they spent practicing in total would be

31/24 + 3/5 + 9/10 = 67/24 =

2 19/24 hours

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In the university library elevator there is a sign indicating a 16-person limit as well as a weight limit of 2750 pounds. Suppos
Alona [7]

Answer:

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

If n variables are added, the mean is n\mu and the standard deviation is s = \sqrt{n}\sigma

In this problem:

n = 16, \mu = 160*16 = 2560, s = \sqrt{16}*27 = 108

What is the probability that the random sample of 16 people in the elevator will exceed the weight limit?

This is 1 subtracted by the pvalue of Z when X = 2750. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{2750 - 2560}{108}

Z = 1.76

Z = 1.76 has a pvalue of 0.961

1 - 0.961 = 0.039

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit

7 0
3 years ago
A sweater that normally costs $32.00 is on sale for $24.00. What is the percent
Vlad1618 [11]

Answer:

25 %

Step-by-step explanation:

Percentage is given by :

\%=\dfrac{\text{original value-new value}}{\text{origial value}}\times 100

Original cost of sweater is $32 and sale cost is $24. So,

\%=\dfrac{32-24}{32}\times 100\\\\=25\%

Hence, 25 % is the markdown on the sweater.

6 0
3 years ago
The solution set of an inequality is shown below.
DENIUS [597]
B is the correct answer because it shows the solution set of an inequality
3 0
3 years ago
!!!!Do not respond with I don't know or anything but help with the answer just for the points!!!!! The average test score A of f
zepelin [54]

Answer:

69

Step-by-step explanation:

8 0
3 years ago
Solve the equation: 3(x - 2) = 36<br><br><br><br><br> X=-5<br> X=14<br> X=10<br> X=5
Paul [167]
3(x-2)=36
3x-6=36
3x=36+6
3x=42
x=42/3
x=14
Answer: 14
I hope this helped mate!
8 0
3 years ago
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