PLEASE HELP ME!!!!!!!!!

What is the value of 4 to the second power - 2 ( 3 times 5 + 1)

valentinak56 [21]

Answer:

4 to the second power = 16. that leaves us with 16 - 2 ( 3 x 5 + 1). next, multiply 3 x 5 = 15. that leaves us with 16 - 2 ( 15 + 1 ). next, add 15 + 1 = 16. that leaves us with 16 - 2 ( 16 ). next, multiply 2 ( 16 ) = 32. finally, subtract 16 - 32 = -16

Step-by-step explanation:

your answer is -16. hope this helps

8 0

3 years ago

In a class of 18 students, 5 are math majors. A group of four students is chosen at random. (Round your answers to four decimal

ss7ja [257]

Answer:

(a) 0.2721

(b) 0.7279

(c) 0.2415

Step-by-step explanation:

(a) If we choose only one student, the probability of being a math major is $\frac{5}{18}$ (because there are 5 math majors in a class of 18 students). So, the probability of not being a math major is $\frac{18}{18} - \frac{5}{18} = \frac{13}{18}$ (we subtract the math majors of the total of students).

But there are 4 students in the group and we need them all to be not math majors. The probability for each one of not being a math major is $\frac{13}{18}$ and we have to multiply them because it happens all at the same time.

P (no math majors in the group) = $\frac{13}{18} *\frac{13}{18}*\frac{13}{18}*\frac{13}{18} = (\frac{13}{18}) ^4$ = 0.2721

(b) If the group has at least one math major, it has one, two, three or four. That's the complement (exactly the opposite) of having no math majors in the group. That means 1 = P (at least one math major) + P (no math major). We calculated this last probability in (a).

So, P (at least one math major) = 1 - P(no math major) = 1 - 0.2721 = 0.7279

(c) In the group of 4, we need exactly 2 math majors and 2 not math majors. As we saw in (a), the probability of having a math major in the group is 5/18 and having a not math major is $\frac{13}{18}$ . We need two of both, that's $\frac{5}{18}*\frac{5}{18}*\frac{13}{18}*\frac{13}{18}$ . But we also need to multiply this by the combinations of getting 2 of 4, that is given by the binomial coefficient $\binom{4}{2}$ .