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2 years ago

Solve the problem 5/n = 1.5/ 1.8

2 answers:

6 0

$\frac{5}{n} = \frac{1.5}{1.8} \ \ \ \Leftrightarrow\ \ \ 5\cdot1.8=1.5\cdot n \ \ \ \Leftrightarrow\ \ \ 9=1.5n\ /:1.5\\\\n=6$

Lostsunrise [7]2 years ago

5 0

5/n=(3/2)/(5/5+4/5)

5/n=(3/2)/(9/5)

5/n=(3/2)*(5/9)

5/n=15/18=5/6

5/n=5/6

30/n=5

5n=30

Therefore: n=6.

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Solve for x. 8(x + 1) - 3(x + 4) = 7(2 - x) Ox=11 Ox=-11 Ox= -1 1 / Ox=11/

inna [77]

Answer:

1.5

Step-by-step explanation:

$8(x + 1) - 3(x + 4) = 7(2 - x) \\ 8 x + 8 - 3x - 12 = 14 - 7x \\ 8x - 3x + 7x = 14 - 8 + 12 \\ 12x = 18 \\ x = \frac{18}{12} \\ x = \frac{3}{2} \\ x = 1.5$

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2 years ago

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What is the mean of the set {4, 6, 8, 10}?

olga2289 [7]

Answer:

Step-by-step explanation:

8 I think

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2 years ago

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The simplified form of 3root45 - root 125 + root 500

bazaltina [42]

Answer:

the answer and explanation is in the picture

please like and Mark as brainliest

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5 0

2 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

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8 0

1 year ago

A sample of 38 students is chosen from a student body where 58% of all students own cell phones. If 24 of the students in this s

Dmitry [639]

Answer:

the sample proportion is higher

Step-by-step explanation:

What we must do is calculate the proportion of the population, we are left with the fact that out of 38 students 58% have cell phones, these are the proportion of the population, then they would be:

38 * 58% = 22.04 students

Now, the sample proportion is 24.

Therefore 22.04 < 24

Which means the sample proportion is higher

8 0

2 years ago

Read 2 more answers