if you had to express a typical NBA player's salary with a single number (round to the nearest 100000) you would use

vova2212 [387]

The second answer.
You have the first prize, x.
Then you have the second prize, which is x - 50.
Then you have the third prize, which is x - 50 -50, or x - 100.
So the total money is:
x + (x-50) + (x-100).

5 0

3 years ago

the question is in the picture below, if you could show all your work so i can see how to do it, that would be appreciated.

ki77a [65]

m∠R = 27.03°

Solution:

Given In ΔQRP, p = 28 km, q = 17 km, r = 15 km

To find the measure of angle R:

Law of cosine formula for ΔQRP:

$r^{2}=p^{2}+q^{2}-2 pq \cos R$

Substitute the given values in the above formula.

$(15)^{2}=(28)^{2}+(17)^{2}-2 (28)(17) \cos R$

$225=784+289-952 \cos R$

$225=1073-952 \cos R$

Switch the given equation.

$1073-952 \cos R=225$

Subtract 1073 from both side of the equation.

$-952 \cos R=-848$

Divide by –952 on both sides.

$$ \cos R=\frac{106}{119}$

$$ R=\cos^{-1} \left(\frac{106}{119}\right)$

$R=27.03^\circ$

Hence m∠R = 27.03°.

7 0

3 years ago

A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec

zhenek [66]

Answer:

b) No, it's not independent.

c) 0.02

d) 0.59

e) 0.57

f) 0.5616

Step-by-step explanation:

To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)

Let's call A the event that she passes the first course, then P(A)=.73

Let's call B the event that she passes the second course, then P(B)=.66

Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98

b) Is the event she passes one course independent of the event that she passes the other course?

Two events are independent when P(A∩B) = P(A) * P(B)

So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:

P(A∪B) = P(A) + P(B) - P(A∩B)

We are going to solve for P (A∩B)

.98 = .73 + .66 - P(A∩B)

P(A∩B) =.73 + .66 - .98

P(A∩B) = .41

Now we will see if the formula for independent events is true

P(A∩B) = P(A) x P(B)

.41 = .73 x .66

.41 ≠.4818

Therefore, these two events are not independent.

c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)

1 - P(A∪B) = 1 - .98 = .02

d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)

1 - P(A∩B) = 1 -.41 = .59

e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.

P(A∪B) - P(A∩B) = .98 - .41 = .57

f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)

P(B|A) = P(A∩B) / P(A)

P(B|A) = .41 / .73 = .5616