Answer:
Yes
Step-by-step explanation:
This is reasonable because if you do :
3/8 - 1/6=
3x3 over 8x3 - 1x4 over 6x4 =
9/24 - 4/24 = 9-4 over 24 =
5/24 and that is greater then 2/5.
Answer:
more than 168 tickets
Step-by-step explanation:
4 weeks = 28 days
5 x 28 = 140
1 week = 7 days
4 x 7 = 28
140 + 28 = 168
In order for hector to sell more than Danielle he has to sell a # of tickets that is greater then 168.
To answer this question, the easier way is to make each of the mixed fractions expressed as improper fractions
Thus, we have 6 4/9 written as 58/9 ( we shall be extending this to the other fractions too; the key is to multiply the denominator by the whole number, then add the numerator)
3 2/9 would be expressed as 29/9
and finally 8 2/9 would be expressed as 74/9
So, we can rewrite the question as follows;
-58/9 - 29/9 - 74/9
We can see all have same denominator;
Thus, we can proceed to summing all the numerators.
We have;
(-58-29-74)/9 = -161/9
We then proceed to turn this to an improper fraction too;
Kindly find the nearest multiple of 9 less than 161, then subtract the value from 161;
That would be 153 and we are left with a difference of 8
So our final mixed fraction would be -17 8/9
Hence; -6,4/9 - 3 2/9 - 8 2/9 = -17 8/9
a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
Answer:
explain
Step-by-step explanation:
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