I need somebody to say they like me so i can beat my friend in a challenge

What is the sum of 22

maxonik [38]

The average of the four numbers is 8

4 0

3 years ago

I’m not that great at these I need help A LOT!! Thanks :)

Art [367]

On the right side of the graphs it shows how many years on the bottom. It ends on te top right which seems to be 50 years.

5 0

3 years ago

Read 2 more answers

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)

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

= $1-(1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}+1140\times 0.20^{3} \times 0.80^{17})$

= 1 - 0.4114 = 0.5886

(d) The expected number of withdrawals is given by;

E(X) = $n\times p$

= $20 \times 0.20$ = 4 withdrawals

3 0

4 years ago

In a mid-size company, the distribution of the number of phone calls answered each day by each of the 12 receptionists is bell-s

iris [78.8K]

Answer:

0.6826

Step-by-step explanation:

Mean(μ) = 37

Standard deviation (σ) = 9

P(28 < x < 46) = ???

Using normal distribution

Z = (x - μ)/σ

For x = 28

Z = (28 - 37)/9

Z = -9/9

Z = -1

For x = 46

Z = (46 - 37)/9

Z = 9/9

Z = 1

We now have

P(-1 < Z < 1)

= P(Z < 1) - P(Z < -1)

From the table, Z = 1 = 0.3413

φ(Z) = 0.3413

Recall that

When Z is positive, P(x<a) = 0.5 +φ(Z)

P(Z<1)= 0.5 + 0.3413

= 0.8413

When Z is negative, P(x<a) = 0.5 - φ(Z)

P(Z< -1)= 0.5 - 0.3413

= 0.1587

We now have

0.8413 - 0.1587

= 0.6826

8 0

3 years ago

If f(x)=4x3+3x2−5x+20 and g(x)=9x3−4x2+10x−55, what is (g−f)(x)?

Montano1993 [528]

Answer:

Step-by-step explanation:

(g-f)(x) = 9x³ - 4x² + 10x - 55 - [ 4x³ +3x² - 5x + 20]

To remove the parenthesis, take - inside, multiply f(x) by -1

= 9x³ - 4x² + 10x - 55 - 4x³ - 3x² + 5x - 20

Now, bring like terms together,

= 9x³ - 4x³ - 4x² - 3x² + 10x + 5x - 55 - 20

= 5x³ - 7x² + 15x - 75

3 0

4 years ago