Let workout Plan A last a hours, and Plan B last b hours.
we are assuming personal training for each client.
i)
"On Monday there were 2 clients who did Plan A and 3 who did Plan"
the total time spent is : 2*a + 3*b =2a+3b
ii)
"On Tuesday there were 4 clients who did Plan A and 8 who did Plan B"
the total time spent was 4*a+8*b=4a+8b
iii) "Joe trained his Monday clients for a total of 7 hours"
so 2a+3b = 7
iv)
"Joe trained his Tuesday clients for a total of 17 hours"
so 4a+8b=17
v) thus we have the following system of equations:
2a+3b = 7
4a+8b=17
multiply the first equation by -2, and then add both equations, to eliminate a:
-4a-6b=-14
4a+8b=17
-------------------
2b=3, so b=3/2
2a+3b = 7
2a+3(3/2)=7
2a+9/2=7
multiply by 2:
4a+9=14
4a=5
a=5/4
Answer :
Plan A lasts 5/4=1.25 h
Plan B lasts 3/2=1.5 h
Answer:
area of rectangle =length ×width
area of rectangle=12yd×15yd
area of rectangle=180yd²
Answer:
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Step-by-step explanation:
Total plants = 11
Domestic plants = 7
Outside the US plants = 4
Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure
Here we have n=4, p=4/11 and q=7/11
P(X≥1) = 1 - P(X<1)
= 1 - P(X=0)
= 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰
= 1 - 0.16399
P(X≥1) = 0.836
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
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