All categories

3 years ago

What is the value of the digit 8 in alaska's population

Mathematics

1 answer:

Montano1993 [528]3 years ago

4 0

Ten thousands is the answer

You might be interested in

g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability th

xxTIMURxx [149]

Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

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.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

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

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

Sampling with replacement:

I consider a success choosing a black ball, so $p = \frac{50}{150+50} = \frac{50}{200} = 0.25$

We want 2 black balls and 2 white, 2 + 2 = 4, so $n = 4$ , and we want P(X = 2).

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

$P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109$

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so $N = 200$

Sample of 4, so $n = 4$

50 are black, so $k = 50$

We want P(X = 2).

$P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116$

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

3 0

2 years ago

Janet has $120. If she saves $20 per week, in how many days will she have $500

Burka [1]

<span>Divide $380 by 20 and then multiply the result by 7. Its really not hard I promise I basically just gave you the answer</span>

4 0

3 years ago

Read 2 more answers

The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party,

Mama L [17]

Answer:

$(a)\ P(x = 0) = 0.2725$

$(b)\ P(x \ge 1) =0.7275$

$(c)\ P(x \le 2) = 0.8948$

Step-by-step explanation:

Given

$n = 8$ --- 8 friends

$p = 15\%$ --- proportion that one-time fling

This question is an illustration of binomial probability, and it is represented as:

$P(X = x) = ^nC_x* p^x * (1 - p)^{n-x}$

Solving (a): P(x = 0) --- None has done one time fling

$P(x = 0) = ^8C_0* (15\%)^0 * (1 - 15\%)^{8-0}$

$P(x = 0) = 1* 1 * (1 - 0.15)^{8}$

$P(x = 0) = 0.85^8$

$P(x = 0) = 0.2725$

Solving (b): $P(x \ge 1)$

To do this, we make use of compliment rule:

$P(x = 0) + P(x \ge 1) =1$

Rewrite as:

$P(x \ge 1) =1 - P(x = 0)$

$P(x \ge 1) =1 - 0.2725$

$P(x \ge 1) =0.7275$

Solving (c): $P(x\le 2)$ --- Not more than 2 has one time fling

This is calculated as:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

We have:

$P(x = 0) = 0.2725$

$P(x = 1) = ^8C_1* (15\%)^1 * (1 - 15\%)^{8-1}$

$P(x = 1) = 8* (0.15) * (1 - 0.15)^7$

$P(x = 1) = 0.3847$

$P(x = 2) = ^8C_2* (15\%)^2 * (1 - 15\%)^{8-2}$

$P(x = 2) = 28* (0.15)^2 * (1-0.15)^6$

$P(x = 2) = 0.2376$

So:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

$P(x \le 2) = 0.2725 + 0.3847 + 0.2376$

$P(x \le 2) = 0.8948$

4 0

3 years ago

X + 2 = 1 what is it

Alexus [3.1K]

-1 as a negative would = X

8 0

3 years ago

Read 2 more answers

1 point) Are the functions f,g, and h given below linearly independent? f(x)=e3x+cos(5x), g(x)=e3x−cos(5x), h(x)=cos(5x). If the

Fynjy0 [20]

Answer:

Functions are linearly dependent (are not linearly independent.)

Step-by-step explanation:

Remember that two functions f(x), g(x) and h(x) are said linearly independent on an interval I if the <em>only solution</em> to the equation

$\alpha f(x)+\beta g(x)+\omega h(x)=0, \ \text{for all } x\in I$

is the trivial one: α = 0, β = 0, ω = 0. If they are not linearly independent, they are called linearly dependent.

Now, let f(x), g(x) and h(x) be the functions:

$f(x)=e^{3x}+\cos(5x),$

$g(x)=e^{3x}-\cos(5x),$

$h(x)=\cos(5x).$

Then, letting α = 1, β= -1 and ω = -2, we see that:

$\alpha f(x)+\beta g(x)+ \omega h(x)=e^{3x}+\cos(5x)-e^{3x}+\cos(5x)+2\cos(5x)=0.$

Hence, the functions f(x), g(x) and h(x) are not linearly independent, or equivalently, are linearly dependent.

8 0

3 years ago