The integers that are less than 2 but greater than -4

If 20% of the people who shop at a local grocery store buy apples, what is the probability that it will take no more than 5 cust

Leokris [45]

Answer:

C) Using a random digits table select one digit numbers where 0-1 is a customer who buys apples and 2-9 is a customer who does not. Keep selecting one digit numbers until you get a 0 or 1. Record the number of digits selected.

Step-by-step explanation:

There are 10 digits from 0 to 10

0-1 are 2 out of 10

2/10 × 100 = 20%

Therefore 20% chance of buying apples

5 0

3 years ago

F(x)=22xplus 33 and g(x)=x2−44x−55, find (g◦f)(88)

Law Incorporation [45]

F(x) = 22x + 33
g(x) = x² - 44x - 55
(g о f)(88) = (22(88) + 33)² - 44(22(88) + 33) - 55
(g о f)(88) = (1936 + 33)² - 44(1936 + 33) - 55
(g о f)(88) = (1969)² - 44(1969) - 55
(g о f)(88) = 3,876,961 - 86,636 - 55
(g о f)(88) = 2,790,325 - 55
(g о f)(88) = 3,790,270

5 0

4 years ago

1.Relection over the y- axis

mart [117]

Answer:

1.

rule={-x,y}

G'={-3,-3}

H'={5,-1}

J'={4,-3}

K'={-2,2}

2.

rule={x,-y}

A'={-4,-12}

B'={0,-10

C'={6,2}

4.=?

5.=?

Step-by-step explanation:

7 0

3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

8 0

4 years ago

Corinna rowed 4/5 mile on a rowing machine in 1/6 hour. What is her rowing rate in miles per hour

DaniilM [7]

Answer: 8.20 miles I believe

6 0

3 years ago