All categories

2 years ago

Pls help me with these 2 questions i will Mark brainlist:( !!

Mathematics

1 answer:

Sav [38]2 years ago

8 0

Answer:

6. b. 50

7. d. 12

Step-by-step explanation:

6. 80% is equal to 0.8.

0.8x =40.

Divide 0.8 by 40 to get x.

x=50.

7. If he needs 3 eggs for 30 cookies, we divide 120 by 30 to get 4.

We multiply 4*3 to find the total amount of eggs needed.

4*3= 12

You might be interested in

2a + 6 = 6<br> 2a =<br> Subtract 6 from both sides<br> Divide both sides by 2

Alla [95]

Answer:

Step-by-step explanation:

2a+6=6

-6=-6

2a=0

2 2

a=0

8 0

3 years ago

3. I think of a number. I subtract 10 and add 5. I then double it. My

ella [17]

Answer:

Step-by-step explanation:

You can divide 50 by 2 which is 25

Then subtract 5

Add 10

And you get your number

It is really simple really ☺️

5 0

2 years ago

In a class there are 10 students who play football and cricket 7 students who do not play football or cricket 14 students who pl

alisha [4.7K]

Answer:

110 members of a sports club play atleast one of the games, football, basketball and volleyball. If 20 play football and basketball only, 15 play football and volleyball only, 26 play basketball and volleyball only, x play all the three games, 2x each play

math

Of 45 students 30 play badminton and 26 play tennis. Each student play either badminton or tennis,determine how many students play:(a)badminton only (b) tennis only

probability and statistics

At a particular school with 200 male students, 58-play football, 40-play basketball, and 8 play both. What is the probability that a randomly selected male student plays a) At least one sport b) Neither sport c) Only basketball

7 0

2 years ago

Graph the following inequality. Click on the graph until the correct one appears.

exis [7]

Answer: Its the first one

You draw the graph of the line y=3x+5

but make the line dotted.

Label the line y>3x+5

Note that if instead we hade y≥3x+5 the line would be solid.

If you draw a line parallel to the y-axis through any x-value the value of y can be any on that vertical line that is above but not on the line y=3x+5

. So you end up with a 'feasible' region of solutions for y

5 0

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

1 year ago