If fh is an angle bisector of triangle efg with

Suppose you have an experiment where you flip a coin three times. You then count the number of heads. a.)State the random variab

WITCHER [35]

Answer:

a) X=number of heads observed when flipped the coin 3 times

b) the probability distribution is

P(X=x) = 3/4 * (1/[(3-x)!*x!)])

or

P(X)=1/8 for x=0 and x=3 and P(X)=3/8 for x=1 and x=2

Step-by-step explanation:

the random variable will be X=number of heads observed when flipped the coin 3 times . Since the result from each flip is independent of the others , then X has a binomial probability distribution , such that

P(X=x)= n!/[(n-x)!*x!)*p^x * (1-p)^(n-x)

where

n= number of times the coin is flipped = 3

p= probability of getting heads in a flip of the coin = 1/2 (assuming that the coin is fair)

therefore

P(X=x)= 3!/[(3-x)!*x!)*(1/2)^(3-x) * (1/2)^x = 3!/[(3-x)!*x!) * (1/2)³ = 3/4 * (1/[(3-x)!*x!)])

P(X=x)= 3/4 * (1/[(3-x)!*x!)]) , for x=[0,1,2,3]

for x=0 and x=3 → P(X)=3/4 * (1/[3!*0!)]) = 1/8

for x=1 and x=2 → P(X)=3/4 * (1/[2!*1!)]) = 3/8

we can verify that is correct since the sum of all the probabilities from x=0 to x=3 is 1/8 + 3/8+ 3/8+ 1/8 = 1

8 0

3 years ago

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Which of the following is a solution of x2 + 2x + 4?

scoray [572]

Answer:8x

Step-by-step explanation:

2++2=4 add x up 4 = 8x

3 0

2 years ago

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Point Q' is the image of Q (4,-6) is under a translation by 1 unit to the right.

alexandr1967 [171]

Answer:

The coordinates of point Q' would be (5, -6) after being translated 1 unit to the right.

Step-by-step explanation:

4 0

3 years ago

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

The area of rectangle is equal to

$A=14*5=70\ in^{2}$

The area of the small semicircle is equal to

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

The area of the larger semicircle is equal to

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

Read 2 more answers

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

Tanzania [10]

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = Number of adults who use their smartphones

The above situation can be represented through the binomial distribution;

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

where, n = number of trials (samples) taken = 15 adult smartphones

r = number of success = at least 4

p = probability of success which in our question is the % of adults

who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}$

= $1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})$

= 0.1773

3 0

3 years ago