These are the steps needed for this question hope this helps if there is any more questions do feel free to ask them
Answer:
1/72=x (part a)
1/70=x (bart b)
Step-by-step explanation:
Divide 1 by 72 for part a, and divide 1 by 70 for part b.
Solve your system of equations.
2x+y=1;4x+2y=−1
Solve 2x+y=1 for y:
2x+y+−2x=1+−2x(Add -2x to both sides)
y=−2x+1
Substitute (−2x+1) for y in 4x+2y=−1:
4x+2y=−1
4x+2(−2x+1)=−1
2=−1(Simplify both sides of the equation)
2+−2=−1+−2(Add -2 to both sides)
0=−3
Answer: No solution. C)
Answer:
Test statistic = -2.44
There is enough evidence to support the strategist's claim.
Step-by-step explanation:
H0 : p = 0.41
H1 : p < 0.41
pˆ = 0.38
Test statistic :
z=pˆ−p/√p(1−p)/n
Z = (0.38 - 0.41) / √(0.41(1 - 0.41) / 1600
Z = - 0.03 / √0.0001511875
Z = - 0.03 / 0.0122958
Z = - 2.4399
Test statistic = -2.44
The Pvalue :
P(Z < -2.44) = 0.0073436
α - level = 0.02
If Pvalue < α ; Reject H0
0.0073436 < 0.02 ; We reject H0
Since Pvalue < α ; Hence, There is enough evidence to support the strategist's claim.
Answer:
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And using this formula we have this:

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917
Step-by-step explanation:
Let X the random variable of interest that a woman must wait for a cab"the amount of time in minutes " and we know that the distribution for this random variable is given by:

And we want to find the following probability:

And for this case we can use the cumulative distribution function given by:
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And using this formula we have this:

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917