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Vera_Pavlovna [14]

3 years ago

12

determine if the following statement is true or false. If false, provide a counter example. If two events are independent, then

the probability of both events is less than one

1 answer:

Mice21 [21]3 years ago

6 0

I'm sorry there is not enough information provided to give an explanation.

You might be interested in

| x-3 | < x-3<br><br> can someone give a step by step process on how to do this?

statuscvo [17]

Answer:

There are no solutions to the inequality.

Step-by-step explanation:

|x - 3| < x – 3

1. Separate the inequality into two separate ones.

(1) x – 3 < x – 3

(2) x – 3 < -(x – 3)

2. Solve each equation separately

(a) Equation (1)

$\begin{array}{rcl}x - 3 & < & x - 3\\x & < & x\\\end{array}\\\text{This is impossible. No solutions exist.}$

(b) Equation (2)

$\begin{array}{rcl}x - 3 & < & -(x - 3)\\x - 3 & < & -x + 3\\x &$

For example, if x = 0, we get

|0 - 3| < 0 - 3 or

3 < -3

4 0

3 years ago

the angle of elevation of the top of an elevator pole to a man standing 10m from it is 60°. find the height of the pole

AleksAgata [21]

Answer:

10√3m

Step-by-step explanation:

Given :-

Angle of elevation = 60°
Distance between man and pole = 60°

We need to find the height of the pole . We are given here base. Therefore using ratio of tan ,

=> tan 60° = p/b

=> √3 = h / 10 m

=> h = 10 m× √3

=> h = 10√3 m

7 0

3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is <u>P(X = 10)</u>, thus:

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

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of <u>x successes on n trials, with p probability</u> of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the <u>p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.</u>

X = 10.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10.5 - 10}{\sqrt{5}}$

$Z = 0.22$

$Z = 0.22$ has a p-value of 0.5871.

X = 9.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{9.5 - 10}{\sqrt{5}}$

$Z = -0.22$

$Z = -0.22$ has a p-value of 0.4129.

0.5871 - 0.4129 = 0.1742.

0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

A similar problem is given at brainly.com/question/24261244

6 0

3 years ago

construct a 90% confidence interval of the population proportion using the giver information x=74 n=150

FrozenT [24]

Answer:

The 90% confidence interval of the population proportion is (0.43, 0.56).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

$CI=\hat p\pm z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

<em>X</em> = 74

<em>n</em> = 150

Confidence level = 90%

Compute the value of sample proportion as follows:

$\hat p=\frac{X}{n}=\frac{74}{150}=0.493$

Compute the critical value of <em>z</em> for 90% confidence level as follows:

$z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

*Use a <em>z</em>-table.

Compute the 90% confidence interval of the population proportion as follows:

$CI=\hat p\pm z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.493\pm 1.645\times \sqrt{\frac{0.493(1-0.493)}{150}}\\\\=0.493\pm 0.0672\\\\=(0.4258,\ 0.5602)\\\\\approx (0.43,\ 0.56)$

Thus, the 90% confidence interval of the population proportion is (0.43, 0.56).

3 0

3 years ago

Craig and Jeff are comparing their plans for downloading songs. Craig pays a monthly fee of $5.50 and then $0.25 per song he dow

Veronika [31]

Answer:

10 songs

Step-by-step explanation:

Let number of songs Craig and Jeff downloaded be s

5.50 + 0.25s = 0.80s

5.50 = 0.55s

s = 10

5 0

3 years ago