Answer:
a. 97.72%
Step-by-step explanation:
The weights of boxes follows normal distribution with mean=28 ounce and standard deviation=0.9 ounces.
a. We have to calculated the percentage of the boxes that weighs more than 26.2 ounces.
Let X be the weight of boxes. We have to find P(X>26.2).
The given mean and Standard deviations are μ=28 and σ=0.9.
P(X>26.2)= P((X-μ/σ )> (26.2-28)/0.9)
P(X>26.2)= P(z> (-1.8/0.9))
P(X>26.2)= P(z>-2)
P(X>26.2)= P(0<z<∞)+P(-2<z<0)
P(-2<z<0) is computed by looking 2.00 in table of areas under the unit normal curve.
P(X>26.2)=0.5+0.4772
P(X>26.2)= 0.9772
Thus, the percent of the boxes weigh more than 26.2 ounces is 97.72%
The combined distance between CD and DE is CE, so...
2x+4 + 6 = x+14 Simplify like terms..
2x + 10 = x + 14 Subtract x from both sides..
x + 10 = 14 Subtract 10 from both sides...
<h2><u><em>
x = 4</em></u></h2>
Proof: 2 (4) + 4 + 6 = 4 + 14
8 + 10 = 18
<em> Check</em>
