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nikdorinn [45]
3 years ago
12

Out of the 32 students in Mr. Martin’s class, 18 are boys. If this ratio is proportional to the ratio of students in the school

and the number of boys in the school, which of the following proportions could could be used to find the number of students in the school? It is given that there are 468 boys in the school.

Mathematics
2 answers:
statuscvo [17]3 years ago
4 0

Answer:

Step-by-step explanation:

kumpel [21]3 years ago
3 0
I believe the answer is D
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Please help solve the question in the image
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Answer:

✅On average Albany gets 20.967 inches rain pee year and Zachary gets 25.153 inches rain per year.

✅On average, it rains more in Zachary in a year.

Step-by-step explanation:

Mean annual rainfall on average = total amount of rainfall for the period of years ÷ 10

✔️Mean Annual rainfall for Albany (in.):

= (22.82 + 19.83 + 28.40 + 23.35 + 30.78 + 21.34 + 11.20 + 17.04 + 18.63 + 16.28) ÷ 10

= 20.967

✔️Mean Annual rainfall for Zachary (in.):

= (26.41 + 23.54 + 30.02 + 26.35 + 34.24 + 25.74 + 18.98 + 21.26 + 23.61 + 21.38) ÷ 10

= 25.153

✅Therefore, on average Albany gets 20.967 inches rain pee year and Zachary gets 25.153 inches rain per year.

✅On average, it rains more in Zachary in a year.

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3 years ago
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3 years ago
Presenting your findings best fits which step of the statistical process?
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Answer:

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3 years ago
Suppose that a box contains 8 cameras and that 4 of them are defective. A sample of 2 cameras is selected at random with replace
Dafna1 [17]

The Expected value of XX is 1.00.

Given that a box contains 8 cameras and that 4 of them are defective and 2 cameras is selected at random with replacement.

The probability distribution of the hypergeometric is as follows:

P(x,N,n,M)=\frac{\left(\begin{array}{l}M\\ x\end{array}\right)\left(\begin{array}{l}N-M\\ n-x\end{array}\right)}{\left(\begin{array}{l} N\\ n\end{array}\right)}

Where x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

The probability distribution for X is obtained as below:

From the given information, let X be a random variable, that denotes the number of defective cameras following hypergeometric distribution.

Here, M = 4, n=2 and N=8

The probability distribution of X is obtained below:

The probability distribution of X is,

P(X=x)=\frac{\left(\begin{array}{l}5\\ x\end{array}\right)\left(\begin{array}{l}8-5\\ 2-x\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}

The probability distribution of X when X=0 is

\begin{aligned}P(X=0)&=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}8-4\\ 2-0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-0)!0!}\right)\times \left(\frac{4!}{(4-2)!2!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end

The probability distribution of X when X=1 is

\begin{aligned}P(X=1)&=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}8-4\\ 2-1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-1)!1!}\right)\times \left(\frac{4!}{(4-1)!1!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.57\end

The probability distribution of X when X=2 is

\begin{aligned}P(X=2)&=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 2-2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-2)!2!}\right)\times \left(\frac{4!}{(4-0)!0!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end

Use E(X)=∑xP(x) to find the expected values of a random variable X.

The expected values of a random variable X is obtained as shown below:

The expected value of X is,

E(X)=∑xP(x-X)

E(X)=[(0×0.21)+(1×0.57)+(2×0.21)]

E(X)=[0+0.57+0.42]

E(X)=0.99≈1

Hence, the binomial probability distribution of XX when X=0 is 0.21, when X=1 is 0.57 and when X=2 is 0.21 and the expected value of XX is 1.00.

Learn about Binomial probability distribution from here brainly.com/question/10559687

#SPJ4

8 0
2 years ago
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