The probability that a randomly selected five-star recruit who chooses one of the best three conferences will be offered a full football scholarship is 69.75%.
<h3>
What is the computation for the above solution?</h3>
Note that this is simple probability.
Hence probability of the 75% football starts selected randomly where the chance is 93% =
75% x 93%
= 69.75%
<h3>
What are the odds a randomly selected five-star recruit will not select a university from one of the three best conferences?</h3>
The chance that 25% of the five-star recruit will not select a university form one of the three best conferences is explained as follows:
Because only 75% stand the chance of being selected into the three ivy leagues, this means that 25% stand no chance.
Recall that the total population is 100%.
<h3>Are these are independent or dependent events. Are they Inclusive or exclusive?</h3>
From the statement of problem given, it is clear that the events are mutually exclusive. Recall "Historically, five-star recruits get full football scholarships 93% of the time, <u>regardless </u>of which conference they go to.
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Answer:
Step-by-step explanation:
We have given:
√2x+3 - √x+1 = 1
First of all isolate the square root of the left hand side:
√2x+3 = √x+1 +1
Now take square on both sides.
(√2x+3)^2 = (√x+1 +1)^2
Open the R.H.S by squaring formula.
∴(a+b)^2 = a^2+2ab+b^2
2x+3 = (√x+1)^2 + 2(√x+1)(1)+(1)^2
2x+3 = x+1 +2√x+1 +1
2x+3 = x+2 +2√x+1
Combine the like terms:
2x-x+3-2 = 2√x+1
x+1 = 2√x+1
Take square on both sides
(x+1)^2 = (2√x+1)^2
x²+2x+1 = 4x+4
x²+2x-4x+1-4 = 0
x²-2x-3 = 0
Now solve the quadratic equation:
a = 1 , b= -2 , c = -3
x = -b+/-√b²-4ac/2a
x = -(-2)+/-√(-2)² - 4(1)(-3) / 2(1)
x = 2 +/- √4+12 / 2
x = 2+/- √16/2
x = 2+/- 4 /2
x = 2+4/2 , x = 2-4/2
x = 6/2 , x = -2/2
x = 3 , x = -1
The solutions we get is (3, -1).
Answer:
a. 5
Step-by-step explanation:
10÷2=5
5÷1=5
15÷3=5
Am bc it is a example as a gh
If x = the fuction hours
and y=kilometers walked
your equation should look like this
y=6.5(2.5)
y=16.25
