We have to define an interval about the mean that contains 75% of the values. This means half of the values will lie above the mean and half of the values lie below the mean.
So, 37.5% of the values will lie above the mean and 37.5% of the values lie below the mean.
In a Z-table, mean is located at the center of the data. So the position of the mean is at 50% of the data. So the position of point 37.5% above the mean will be located at 50 + 37.5 = 87.5% of the overall data
Similarly position of the point 37.5% below the mean will be located at
50 - 37.5% = 12.5% of the overall data
From the z table, we can find the z value for both these points. 12.5% converted to z score is -1.15 and 87.5% converted to z score is 1.15.
Using these z scores, we can find the values which contain 75% of the values about the mean.
z score of -1.15 means 1.15 standard deviations below the mean. So this value comes out to be:
150 - 1.15(25) = 121.25
z score of 1.15 means 1.15 standard deviations above the mean. So this value comes out to be:
150 + 1.15(25) = 178.75
So, the interval from 121.25 to 178.75 contains the 75% of the data values.
Answer:
0
Step-by-step explanation:
x and x are same terms.
Answer:
It comes with 13.8 ounces of crackers now.
Step-by-step explanation:
12 + 15% = 13.8
There are 5 solutions for this system.
x^2 + 4y^2 = 100 ____1
4y - x^2 = -20 ____2
Add both 1 & 2 together. x^2 gets cancelled
4y^2 + 4y = 80 (send 80 to the other side and divide by 4)
Then equation the becomes : y^2 + y -20 =0
Now factorise the equation: (y+5) (y-4) = 0
Solve for y : y = -5 and y = 4
Using the values of y to find the values of x. From equation 1:
x^2 = 100 - 4y^2 x = /100 - 4y^2 (/ means square root) Replace values of y
y = -5, x = /100 - 4(-5)^2 = /100 - 100 = 0
y = 4, x = /100 - 4(4)^2 = / 100 - 64 = /36 = -6 or 6
Thus we have 6 solutions y = -5, 4 and x = -6, 0, 6
Answer:
The value of x is 30°
Step-by-step explanation:
X° + 90° = 129° (Exterior angle property)
x° = 129° - 99° = 30°
<u>x°</u><u> </u><u>=</u><u> </u><u>3</u><u>0</u><u>°</u>
Hope it helps:)