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2 years ago

Help please me!!!!!!!!!!!

Mathematics

2 answers:

mr_godi [17]2 years ago

7 0

To find area, you do length times width. In this case, 5 inches is both the length and width, so you do 5 times 5 and you get 25.

The answer is 25 in²

AlexFokin [52]2 years ago

5 0

D: We solve this in exactly the same way in which we solved the previous area problem. Side length is s, area is s^2. Here, side length is 5 in; area is 25 in^2.
the problem does say, however, not to include units in your answer. Thus, just write "25."

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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

Find the difference between the square of 18 and the square of 9

NemiM [27]

Answer:

18 x 9 = 162

Step-by-step explanation:

7 0

2 years ago

HELP ME POR FAVOR PLEASE PLEASEE

Basile [38]

Answer:

$x=8\frac34\ \,,\quad y=8\frac12$

Step-by-step explanation:

$\Delta TRS\sim \Delta LJK \quad\implies\quad \dfrac{|TR|}{|LJ|}=\dfrac{|RS|}{|JK|}=\dfrac{|ST|}{|KL|}\\\\\dfrac{4}{7}=\dfrac{5}{x}=\dfrac{6}{y+2}\\\\\\\dfrac{4}{7}=\dfrac{5}{x}\quad\implies\quad4x=5\cdot7\quad\implies\quad x=\dfrac{35}4=8\frac34\\\\\\\dfrac{5}{x}=\dfrac{6}{y+2}\quad\implies\quad 5(y+2)=6\cdot\dfrac{35}4\\\\5y+10=3\cdot\dfrac{35}2\\\\5y\ =\ \dfrac{3\cdot35}2-10\\ \div5\qquad\quad \div5\\\\y=\dfrac{3\cdot7}2-2=10\frac12-2=8\frac12$