Answer:


Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

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.
Middle 85%.
Values of X when Z has a pvalue of 0.5 - 0.85/2 = 0.075 to 0.5 + 0.85/2 = 0.925
Above the interval (8,14)
This means that when Z has a pvalue of 0.075, X = 8. So when
. So




Also, when X = 14, Z has a pvalue of 0.925, so when 




Replacing in the first equation





Standard deviation:




Yes it is, and don't worry, I hate school too! Amazing username!
Answer:
from my understanding te correct time will be at 10:15am
Step-by-step explanation:
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>
As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.
Assumption : Let us assume the length as "l" and width as "b". So,
Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.
- <em>l</em> denotes length
- <em>b</em> denotes breadth


Now, finding the length. According to the question,

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>
Answer:
15( b+ a) - 2(a + 0 ) = c
then
c+b = t
We could put the two together to make 15(b+a)- 2(a+0) + b = t
Where t = 8379.70
a= £249.99 b= 329.99
Step-by-step explanation:
Equal amounts are 8549.7 from 30 mowers a+b = small and large.
If we use this as an even and take away a+b at different equations we can find the answer to be 8379.70 on the 2nd equation. First equation = 8579.70
8549.7- 249.99-249.99+329.99 to equal the given year to find c in first equation and total in year = t within the above equation.