Explain and answer please throughlly explain. I got 7 with a remainder of 3 halp. I think its .73 idk.
1 answer:
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Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60
Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.
The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>Another way to solve this is using substitution for one of the radicals. We choose ...
Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.
