Create a list of six values where the mean, median, and mode are 45, and only two of the values are the same.

Ymorist [56]

AB=YZ
AC=YX
CB=XZ
it’s because if you look at the lengths of the figure the longer lines are similar and the shorter lines are similar AB was the shortest line on the bigger triangle and YZ was the shortest on the smaller triangle then all you had to do was figure out which side was the bottom line then you’d have all your answers

6 0

3 years ago

Divide the fractions: 2 / 5 ÷ 3 / 4.

Serjik [45]

Answer:

8/15

Step-by-step explanation:

KEEP FLIP FLIP

keep 2/5

flip the sign (multiply, not divide)

flip 3/4 to 4/3

2/5 x 4/3

8/15

4 0

3 years ago

Helppp please please helpppp

Anna71 [15]

ANSWER is 5v+2b=117
Explanation i just know

7 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

Mkey [24]

Answer:

(a) Probability that 2 or fewer will withdraw is 0.2061.

(b) Probability that exactly 4 will withdraw is 0.2182.

(c) Probability that more than 3 will withdraw is 0.5886.

(d) The expected number of withdrawals is 4.

Step-by-step explanation:

We are given that a university found that 20% of its students withdraw without completing the introductory statistics course.

Assume that 20 students registered for the course.

The above situation can be represented through binomial distribution;

$P(X =r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,......$

where, n = number of trials (samples) taken = 20 students

r = number of success

p = probability of success which in our question is probability

that students withdraw without completing the introductory

statistics course, i.e; p = 20%

Let X = Number of students withdraw without completing the introductory statistics course

So, X ~ Binom(n = 20 , p = 0.20)

(a) Probability that 2 or fewer will withdraw is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}$

= $1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}$

= 0.2061

(b) Probability that exactly 4 will withdraw is given by = P(X = 4)

P(X = 4) = $\binom{20}{4} \times 0.20^{4} \times (1-0.20)^{20-4}$

= $4845\times 0.20^{4} \times 0.80^{16}$

= 0.2182

(c) Probability that more than 3 will withdraw is given by = P(X > 3)

P(X > 3) = 1 - P(X $\leq$ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1-(\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}+\binom{20}{3} \times 0.20^{3} \times (1-0.20)^{20-3})$