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>First, you need to take the 15% off from the $120 so it will make it easier for you to calculate later. Since there is 15% off, your $120 becomes only $102. Since your keyboards only costs $102 each now, you just have to multiply the number of keyboards you wanted to the less price. $102 multiplied by 24 keyboards is equals to $2,448. We don’t have to add taxes because it is stated that we are just looking for SUBTOTAL before taxes. The answer is A. $2,448. <span>
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Answer (2x + 7) • (x2 - 2)
Step-by-step explanation:
please give me brainlest!!
Step by step solution :
Step 1 :
Trying to factor as a Difference of Squares :
1.1 Factoring: x2-2
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Final result :
(2x + 7) • (x2 - 2)
A. Definition of Congruence
