Answer: B I believe
Step-by-step explanation:
Answer: big boi like a someboody 2x -853
Step-by-step explanation:
<span> $843.44 is the answer to this problem
</span>
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!!!
<span>So we want to know how much ball bearings can be made with 5.24 cm^3 if one ball bearing has a diameter of 1 cm. We know that radius r=d/2=0.5cm So the volume V of one ball bearing is: V=(4/3)*pi*r^3 so V=(4/3)*3.14*(0.5)^3cm^3=0.524cm^3. Now we simply divide the volume of steel by the volume of the ball bearing: 5.24/0.524=10. So we can make 10 ball bearings from 5.24 cm^3 of steel</span>
