Your answer would be 4.
If we substitute in x = -8 and y = -4 into the expression, we get:
This then becomes:
= 
= 4.
I hope this helps!
Answer:
Use 49 ounces of the 14% allow and 41 ounces of the 23% alloy.
Step-by-step explanation:
Each ounce of the 14% copper contains 0.14 ounce of pure copper.
Each ounce of the 23% copper contains 0.23 ounce of pure copper.
Each ounce of the 18.1% copper contains 0.181 ounce of pure copper.
Use x ounces of the 14% and y ounces of the 23% to make 90 ounces of 18.1% alloy.
x+y = 90
y = 90-x
0.14x + 0.23y = 0.181·90
0.14x + 0.23(90-x) = 16.29
0.14x + 20.7 - 0.23x = 16.29
-0.09x + 20.7 = 16.29
4.41 = 0.09x
x = 49
y = 90-x = 41
Use 49 ounces of the 14% allow and 41 ounces of the 23% alloy.
Answer:
3/8
Step-by-step explanation:
-3/2 can change to -12/8
7/4 can change to 14/8
1/8 stays the same
-12/8 + 14/8 = 2/8 + 1/8 = 3/8
Answer:
0.347% of the total tires will be rejected as underweight.
Step-by-step explanation:
For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.
And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.
1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344
1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792
The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)
Using data from the normal distribution table
P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight
Hope this Helps!!!
