1 is a, since the MN would match up with PQ and PR would match up with MO
2 7x - 5 = 5x + 11
2x - 5 = 11
2x = 16
x = 8
Plug x back in
7(8) - 5 = 56 - 5 = 51 so B
Answer:
For a uniform move, the traveled distance equation is
y = v*x
where "v" is the average speed (or rate), in miles per minute, x is the time, in minutes.
To find "v" from the given data, divide 455 miles by 65 minutes (the flight time between the cities)
v = 455%2F65 = 7 miles per minute.
So, finally, your equation is
y = 7x.
Key word here, "or"...so you would add the unique probabilities of either together...
P(e)=1/2, P(>4)=3/10 (the numerator is three rather than five because 2 and 4 were already accounted for in P(e)...
P=5/10+3/10=8/10=4/5 (80%)
P=0.8
Answer:
The probability of getting someone who tests positive, given that he or she had the disease is 0.8954.
Step-by-step explanation:
The data provided is:
YES NO Total
Positive 137 8 145
Negative 16 139 155
Total 153 147 300
An individual is selected at random from the group.
We need to compute the probability of getting someone who tests positive, given that he or she had the disease.
The conditional probability of an event <em>A</em> given that another event <em>B</em> has already occurred is:

Let <em>A</em> = individuals who tests positive and <em>B</em> = individual who had the disease.
The number of individuals who tests positive and had the disease is,
n (A ∩ B) = 137
n (B) = 153
Compute the conditional probability of A given B as follows:



Thus, the probability of getting someone who tests positive, given that he or she had the disease is 0.8954.