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2 years ago

Solve PLEASE HELP AND HURRY!

Mathematics

1 answer:

Luda [366]2 years ago

3 0

Answer:

it's -7 1/3 because since the equation is all negative you would just simply add the -5 and -2 1/3

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A chess player played 44 games total. If he won 16 of the games, what is the ratio of games he lost to games he won?

BlackZzzverrR [31]

56 ratio is it so yep

7 0

4 years ago

PLEASE HELP !!!!!!!!!! WILL FOLLOW

antoniya [11.8K]

Answer:

1/12

Step-by-step explanation:

2/3=8/12

2x4=8

3x4=12

1/4=3/12

1x3=3

4x3=12

8/12+3/12=11/12

1-11/12= 1/12

8 0

3 years ago

Answer the question please:<br> The correct answer will be marked as brainliest =)<br> thanks xoxo

Nataly [62]

100,000,000 is the answer

4 0

3 years ago

Read 2 more answers

A filtration process removes a random proportion of particulates in water to which it is applied. Suppose that a sample of water

Doss [256]

Answer:

$E(Y)=\frac{1}{25}$

Step-by-step explanation:

Let's start defining the random variables for this exercise :

$X_{1}:$ '' The proportion of the particulates that are removed by the first pass ''

$X_{2}:$ '' The proportion of what remains after the first pass that is removed by the second pass ''

$Y:$ '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

$Y=(1-X_{1})(1-X_{2})$

We also assume that $X_{1}$ and $X_{2}$ are independent random variables with common pdf.

The probability density function for both variables is $f(x)=4x^{3}$ for $0$ and $f(x)=0$ otherwise.

The first step to solve this exercise is to find the expected value for $X_{1}$ and $X_{2}$ .

Because the variables have the same pdf we write :

$E(X_{1})= E(X_{2})=E(X)$

Using the pdf to calculate the expected value we write :

$E(X)=\int\limits^a_b {xf(x)} \, dx$

Where $a=$ ∞ and $b=$ - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between $0$ and $1$ ⇒

Using the pdf we calculate the expected value :

$E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}$

⇒ $E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}$

Now we need to use some expected value properties in the expression of $Y$ ⇒

$Y=(1-X_{1})(1-X_{2})$ ⇒

$Y=1-X_{2}-X_{1}+X_{1}X_{2}$

Applying the expected value properties (linearity and expected value of a constant) ⇒

$E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})$

Using that $X_{1}$ and $X_{2}$ have the same expected value $E(X)$ and given that $X_{1}$ and $X_{2}$ are independent random variables we can write $E(X_{1}X_{2})=E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-2E(X)+[E(X)]^{2}$

Using the value of $E(X)$ calculated :

$E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}$

$E(Y)=\frac{1}{25}$

We find that the expected value of the variable $Y$ is $E(Y)=\frac{1}{25}$

3 0

3 years ago

Given data set:

Murrr4er [49]

Answer:

mean =6, median =6.5 mode = 7

Step-by-step explanation:

mean =6, median =6.5 mode = 7

hello,

to find the mean, you add all the numbers together and divide it by the total number of numbers. so that would be 66(total sum) divided by 11 (number of numbers) to get 6.

to find the median, you would organize the numbers from least to greatest and find the middle number which is 6.5

last but not least, to find the mode, just look for the number that appears most often. Therefore, the mode would be 7.

hope this helped!

7 0

3 years ago