All angle measure is preserved, therefore measure of A' is 115
Answer:
The parcel with weight less than 20.14 pounds are 99% of all parcels under the surcharge weight.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12 pounds
Standard Deviation, σ = 3.5 pounds
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.99
Calculation the value from standard normal z table, we have,
Thus, parcel with weight less than 20.14 pounds are 99% of all parcels under the surcharge weight.
The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.
Find the z score for 93.8 inches:
93.8 inches - 91.0 inches 2.8 inches
z = ------------------------------------- = ----------------- = 1.4
2 inches 2 inches
Now find the area under the standard normal curve to the left of z = +1.4.
My calculator returns the following:
normalcdf(-100,1.4) = 0.919. This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.
Y=5-2x
x^2 + (5-2x)^2 = 25
x^2 + 25-20x+4x^2 = 25
x^2 +4x^2-20x+25=25
5x^2 -20x=0
x=0 and y=5
Using row 4:
<span>coefficients are: 1, 4, 6, 4, 1 </span>
<span>a^4 + a^3b + a^2b^2 + ab^3 + b^4 </span>
<span>Now adding the coefficients: </span>
<span>1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4 </span>
<span>Substitute a and b: </span>
<span>a = 4x </span>
<span>
b = -3y </span>
<span>1(4x)^4 + 4(4x)^3(-3y) + 6(4x)^2(-3y)^2 + 4(4x)(-3y)^3 + 1(-3y)^4 </span>
<span>Now simplify the above: </span>
<span>256x^4 - 768x^3y + 864x^2y^2 - 432xy^3 + 81y^4 </span>
