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3 years ago

10

Jordan Has 9 Blue Balls, 6 Red Balls And 5 Brown Balls In A Box. Two Balls Are Drawn Without Replacement From The Box. What Is T

he Probability That Both Of The Balls Are Brown?

2 answers:

marshall27 [118]3 years ago

5 0

5/20 .
add up the number of all the balls combined
5 out of the 20 total are brown

Hunter-Best [27]3 years ago

4 0

Well the chances of one being brown would be 5/20, so i believe the chances of getting two brown would be 1/16

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An account earns simple interest.

olasank [31]

Data:
P (Principal) = $1500
I (Simple Interest) = ?
t (time) = 5 years
r (percentage rate) = 4% = 0.04

Formula:
$I = P*r*t$

Solving (1) <span>Find the interest earned.</span>
$I = P*r*t$
$I = 1500*0.04*5$
$\boxed{\boxed{I = 300}}\end{array}}\qquad\quad\checkmark$

Solution (2) <span>Find the balance of the account.
</span>The future value, (Account), of a loan is given by the equation: A = P + I
$A = P + I$
$A = 1500 + 300$
$\boxed{\boxed{A = 1800}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

We learned in Exercise 4.18 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample o

liraira [26]

Answer:

a) 108 people with a standard deviation of 3.286335

b) No

c) 0.218163 or around 21.82%

d) See explanation below.

Step-by-step explanation:

This situation can be modeled with the Binomial Distribution which gives t<em>he probability of an event that occurs exactly k times out of n, and is given by </em>

<em> $\large P(k;n)=\binom{n}{k}p^kq^{n-k}$ </em>

<em>where </em>

<em> $\large \binom{n}{k}$ = combination of n elements taken k at a time. </em>

<em>p = probability that the event (“success”) occurs once </em>

<em>q = 1-p </em>

In this case, the event “success” is finding an American adult who had chickenpox before adulthood with probability p=0.9

and n=120 American adults in the sample.

The <em>standard deviation for this binomial distribution</em> is

$\large s=\sqrt{npq}$

where <em>n is the sample size 120 </em>

a)

We consider a random sample of 120 American adults. How many people in this sample would you expect to have had chickenpox in their childhood?

If about 90% of American adults had chickenpox before adulthood, we expect to find 90% of 120 = 0.9*120=108 people in the sample who had chickenpox in their childhood.

The standard deviation would be

$\large s=\sqrt{120*0.9*0.1=3.286335}$

b)

Would you be surprised if there were 105 people who have had chickenpox in their childhood?

No, because 105 and 108 are in the interval [mean - s, mean +s]

c)

What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

The probability that 105 or fewer people in this sample have had chickenpox in their childhood is

$\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}$

We compute this value easily with a spreadsheet and we get

$\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}=0.218163\approx 21.82\%$

d)

How does this probability relate to your answer to part (b)?

A binomial distribution with np>5 and nq>5 with n the sample size as is the case here, behaves pretty much like a Normal distribution with mean np and standard deviation $\large \sqrt{npq}$ , so around 60% of the data are in the interval [mean -s, mean +s] and 40% outside, so roughly <em>20% of the data should be in [0, mean-s] </em>

4 0

3 years ago

Two friends are hiking in Death Valley National Park. Their elevation ranges from 228 ft below sea level at

monitta

Answer:

x - 231 ≤ 459

Step-by-step explanation:

Given:

Elevation ranges below sea level = 228 ft

Elevation ranges above sea level = 690 ft

Elevation = x

Computation:

Ideal range = [690-228] / 2 = 231

Tolerance range = [690+228] / 2 = 459

So,

x - 231 ≤ 459

6 0

3 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. 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.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

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

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

So

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago

In the Olympics, the athlete received an 85% score on the compulsory exercises, which are weighted 60%. In the optional round, w

mote1985 [20]

Answer:

jjhbnjycfgkjjhvgyh hv bv h. bgg h. hb.

4 0

3 years ago