Answer:
The short answer is there isn’t.
Start by writing each of these as an expression:
x * y = 60
x + y = 7
Next, solve each for the same variable; in this case, y:
(x * y) / x = 60 / x
.: y = 60 / x
(x + y) - x = 7 - x
.: y = 7 - x
Next, replace y of the second expression to the first
y = 60 / x & y = 7 - x
.: 7 - x = 60 / x
Now, solve for x:
(7 - x) * x = (60 / x) * x
.: x * 7 - x^2 = 60
This is quadratic, so write it in the form of ax2 + bx + x = 0
(-1)x^2 + (7)x + (-60) = 0
.: a = -1, b = 7, c = -60
Finally solve for b:
x = (-b +- sqrt(b^2 - 4*a*c)) / 2a
.: x = (-7 +- sqrt(7^2 - 4*-1*-60)) / (2 * -1)
.: x = (-7 +- sqrt(49 - 240)) / -2
.: x = (-7 +- sqrt(-191)) / -2
The square root of a negative value is imaginary and thus there’s no real answer to this problem.
I’m going to say B I could be wrong
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!!!
The answer is 0.0017. Do you not have a calculator?
