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!!!
3y + 12 = 6x
3y = 6x - 12
y = 2x - 4
2y = 4x + (-8)
y = 2x + (-4)....the same as y = 2x - 4
b would have to be a -8
Answer:
pues que es en que necesitas ayuda
Step-by-step explanation:
dime namas comentame y te ayudo
3x^2 + 8x - 4 - (6x^2 - 5x + 3)....distribute thru the parenthesis
3x^2 + 8x - 4 - 6x^2 + 5x - 3 .....combine like terms
-3x^2 + 13x - 7 <==
Answer:
<h2>
∠PZQ = 63°</h2>
Step-by-step explanation:
If point P is the interior of ∠OZQ , then the mathematical operation is true;
∠OZP + ∠PZQ = ∠OZQ
Given parameters
∠OZQ = 125°
∠OZP = 62°
Required
∠PZQ
TO get ∠PZQ, we will substitute the given parameters into the expression above as shown
∠OZP + ∠PZQ = ∠OZQ
62° + ∠PZQ = 125°
subtract 62° from both sides
62° + ∠PZQ - 62° = 125° - 62°
∠PZQ = 125° - 62°
∠PZQ = 63°
<em>Hence the value of ∠PZQ is 63°</em>
