The last graph is the answer
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.
Answer:
The customary system states that one pound is equivalent to sixteen ounces(oz).
Step-by-step explanation:
Since the package is one pound, it contains sixteen ounces(oz).
Any time you have a fraction within an equation, multiply the entire equation by the denominator to clear the fraction. Since the lead term is negative, we can multiply that away as well
(-14) (0=-1/14x²+4x+5) [now distribute]
0=x²-56x+70 [try to factor into binomials first]
Since 70 only has prime factors of 2·5·7, there is no combination which equals (-56). Use the quadratic formula, or complete the square. I'll use quadratic:
x=<u>-b+/-√(b²-4ac)</u>
2a
a=1, b=(-56), c=70
x= <u>-(-56)+/- √((-56)²-4(1)(70)</u>
2(1)
x= <u>56+/- √(3136-280)
</u> 2
<u />x=<u>56+/-√(2856)</u>
2
x=<u>56+/-√(2³·3·119)</u>
2
x=<u>56+/-2√(714)</u>
2
x=28+√714; x=28-√714