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!!!
Did you ever find the answer?
2X^2+22x+318=1270
2x^2+22x-952=0
x^2+11x-476=0
(x+28)(x-17)=0
x=17
I divided the two fractions & drew a model.

So we know that the bird seed will last 2.5 days.
Knowing this, we can go with:
B. Rectangle model divided into six equal sections, five sections are labeled five-sixths and two sections are labeled one-third, equaling two and one-half days.
Let me know if you find any problems with my answer.