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Answer:
P [ 55 ≤ X ≤ 70 ] = 23.55 %
Step-by-step explanation:
We know:
μ₀ = mean of population σ = standard deviation
We begin for calculating the probability of X ≤ 55 so
Step 1:
μ₀ = 50 σ = 7
×₁ = ( μ - μ₀ ) ÷ σ ⇒ ×₁ = (55-50) ÷ 7 ⇒ ×₁ = 5 ÷ 7 ×₁ = 0.7143
We look z table and we have to interpolate
value (closest smaller Than ×₁ 0.71 and closest bigger than ×₁ 0.72
The associated area for these point are: 0.7611 and 0.7642
Taking diferences and by rule of three
0.01 ⇒ 0.0031
0.0042 ⇒ Δ Δ = 0.0013 and Area for ×₁ = 0.7624
That is the probability of X ≤ 55
Now we have to find the probability of X ≤ 70
We procede as in step 1
μ₀ = 50 σ = 7
×₂ = ( μ - μ₀ ) ÷ σ ⇒ ×₂ = (70-50) ÷ 7 ⇒ ×₂ = 20 ÷ 7 ×₂ = 2.8571
We look z table and we have to interpolate
for 2.85 ⇒ 0.9978 (area) 2.8571 - 2.85 = 0.007
for 2.86 ⇒ 0.9979 (area)
Therefore by rule of three we have
0.01 ⇒ 0.0001
0.007 ⇒ Δ Δ = 0.00007
And area for ×₂ = 2.8571 is equal to 0.99787 and the probability of
X ≤ 0.99787 or 99.79
Now if we look the annex drawing (the area we are looking for is enclose for the values x₁ and ×₂ (pink area) and is the diference between these two areas
So P [ 55 ≤ X ≤ 70 ] is 99.79 % - 76.24 % = 23.55
Answer:
Total surface area is 630cm
Step-by-step explanation:
Surface area = (Perimeter of the base × Length) + (2 × Base Area) = (S1 +S2 + S3)L + bh
Let A, B and C be the amount machines A, B and C, respectively, produce each hour alone.
With all three machines together produce 520 drinks each hour, we know that the addition of the amount that each produces each hour is equal to 520, that is.
Now, similarly, machines A and B produce together 320 drinks each hour, so the sum of the amounts A and B produce each hour is equal to 320, so:
If machine C produce 30 more drinks than B each hour, than the amount of B plus 30 will be the amount of C:
So, we got the system of equations:
To solve, notice that A + B is known, because of the second equation, so we can substitute A + B by 320 in the first equation:
Now, we can substitute C into the third equation and solve for B:
And, finally, we can substitute B into the second equation and solve for A:
Answers:
First box:
Second box:
Third box:
Fourth box: