All categories

3 years ago

For a fundraiser , Brad sold 2 less than 3 times the number of candy bars calls that Mary did . If they sold 334 total bars, fin

d the number sold by each . Show your work

Mathematics

1 answer:

marysya [2.9K]3 years ago

8 0

$x-\ Mary\\
3x-2-\ Brad\\\\
x+3x-2=334\\
4x-2=334\ \ |add\ 2\\
4x=336\ \ | divide\ by\ 4\\x=84\\
3x-2=3*84-2=252-2=250\\\\
Brad\ sold\ 250\ bars\ and\ Mary\ sold\ 84\ bars.$

You might be interested in

Plz help asap for brainiest

Sergio [31]

Answer: x = 16

Step-by-step explanation:

If RQ is the bisector, that means the angle is cut in half. Therefore angles PRQ and QRS are equal to each other.

4x+6 = 7x-42

subtract 7x from both sides

-3x+6 = -42

subtract 6 from both sides

-3x = -48

divide both sides by -3

x = 16

7 0

2 years ago

Read 2 more answers

3 times (-8) times (-2)

dexar [7]

Answer:

Step-by-step explanation:

Signals during a multiplication of two numbers:

They have the same signal: The multiplication is positive.

They have different signals: The multiplication is negative.

3 times (-8) times (-2)

First 3*(-8):

Different signals, so negative. 3*8 = 24. So the answer is -24.

Then:

3*(-8)*(-2) = (-24)*(-2)

Same signal, so positive. 24*2 = 48. The answer is 48.

4 0

1 year ago

Plz answer only if u know

Flauer [41]

The slope would be -5

3 0

2 years ago

Read 2 more answers

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

What's the answer for this? CLASS 6 MATHS I NEED HELP ASAP!!!

devlian [24]

Answer:

20:

3 5/12 or 41/12

21:

-9 5/8 or -77/8

7 0

2 years ago