The median of the five numbers in the box is 10 the range is 12
1 answer:
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So
if you assume that the month has 30 days and that the library opens at midnight, then
24 hours in a day
5 pm=12+5=17 hours
on wednessday=17-2=15 hours
wednessday=1/7 of week
so we find 1/7 of 30 which is 30/7=4 and 2/7
then subtract that from 30
30-4 and 2/7=25 and 5/7
ok so then we have
25 and 5/7 days is 17 hours and
4 and 2/7 days is 15 hours
so just multipy them and add
25 and 5/7 times 17=437.143 hours
4 and 2/7 days times 15 =64.2857
add
437.143+64.2856=501.429
so aprox 501.429
the real equation is![[(\frac{1}{7})(n)(15)]+[ (\frac{6}{7}) (n)(17)]=hours](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B7%7D%29%28n%29%2815%29%5D%2B%5B%20%28%5Cfrac%7B6%7D%7B7%7D%29%20%28n%29%2817%29%5D%3Dhours)
where n represents the number of days in the month
apros 501.429
if you assume that the month has 30 days and that the library opens at midnight, then
24 hours in a day
5 pm=12+5=17 hours
on wednessday=17-2=15 hours
wednessday=1/7 of week
so we find 1/7 of 30 which is 30/7=4 and 2/7
then subtract that from 30
30-4 and 2/7=25 and 5/7
ok so then we have
25 and 5/7 days is 17 hours and
4 and 2/7 days is 15 hours
so just multipy them and add
25 and 5/7 times 17=437.143 hours
4 and 2/7 days times 15 =64.2857
add
437.143+64.2856=501.429
so aprox 501.429
the real equation is
apros 501.429
Answer:
for brand x for 1 fluid ounces
for brand y
0,75-0,65=0,1
Brand x costs $0,1 less
Step-by-step explanation:
Answer:
The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Population:
Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000
Sample of 40
Shape approximately normal
Mean 350000
Standard deviation
The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.