Ask question

Ask question

All categories

3 years ago

9

according to a survey conducted by the Texas Free Press 25 out of 500 people do not have internet access at their home what perc

ent of the people surveyed to have internet access at their home

1 answer:

guapka [62]3 years ago

4 0

Answer:

95%

25/500=.05

1 -.05=.95

You might be interested in

A bond has a $12,000 face value, a 9-year maturity, and a 3.1% coupon. Find the total of the interest payments paid to the bondh

jeka57 [31]

Answer:

$15348

Step-by-step explanation:

i = prt is the simple interest equation

i = 12,000(0.031) = 372

i = 372*9 = 3348

12000+3348 = $15348

4 0

2 years ago

when multiplie discount are used if an item is first discount by 40%,then again by 30%.then again by 10%,the final or cumulative

nadezda [96]

The answer is : 40+30=70
70+10=80 therfore the final or cumulative discount is 80%

4 0

3 years ago

Michael needs to cut a rod 11-4 long.which point shows 11-4

Novay_Z [31]

S because 11/4 is equivalent to 2 3/4

6 0

4 years ago

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of flights i

Morgarella [4.7K]

Answer:

a) $P(X = n) = (0.85)^{n}$

b) $P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

c)

$P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)$

For each value, we use

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

d) If $x > E(X) + \sqrt{V(X)}$ , yes, otherwise no

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrived on time, or it did not. Flights are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A measure is said to unusual if it is more than 2.5 standard deviations from the mean.

85% of recent flights have arrived on time.

This means that $p = 0.85$

(a) Find the probability that all of the flights were on time.

This is $P(X = n)$ , in which n is the size of the sample studied.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = n) = C_{n,n}.(0.85)^{n}.(0.15)^{n-n}$

$P(X = n) = (0.85)^{n}$

(b) Find the probability that exactly x of the flights were on time.

Exactly x of the flights: So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

(c) Find the probability that x or more of the flights were on time.

This is

$P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)$

For each value, we use

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

(d) Would it be unusual for x or more of the flights to be on time?

If $x > E(X) + \sqrt{V(X)}$ , yes, otherwise no

8 0

3 years ago

What is the sum of the first 7 terms for the geometric sequence with first term 3, seventh term 46,875, and common ratio 5

olga55 [171]

Have a nice day! Hopefully this could help you

5 0

3 years ago

Read 2 more answers