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3 years ago

If Sally had six balls and Mark had seven how many balls did he have in all

Mathematics

2 answers:

kari74 [83]3 years ago

8 0

Step-by-step explanation:

he had 7 balls

hope he healthy that's a lot of balls to have

AURORKA [14]3 years ago

4 0

Answer:

13 i think 6+7 = 13 yeh yeh

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Helppppppp pleaseeeeee

ANEK [815]

All you have to do is plug in the number for x, and find y. I will write this out as if in that chart you have shown.

x y = 9 - 3x y

_______________________________________

0 y = 9 - 3(0) 9

1 y = 9 - 3(1) 6

2 y = 9 - 3(2) 3

3 y = 9 - 3(3) 0

Hope that helps!!

4 0

3 years ago

Read 2 more answers

Please help me with this sum

barxatty [35]

Answers:

781 × 8 = 6248

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8 0

2 years ago

Ten of 16 students in Nyack's class are girls. His teacher selected two helpers by randomly drawing names. He drew a boys name f

katrin [286]

$|\Omega|=16\cdot15=240\\|A|=6\cdot10=60\\\\P(A)=\dfrac{60}{240}=\dfrac{1}{4}$

He's wrong.

7 0

3 years ago

A random variable X with a probability density function () = {^-x > 0

Sliva [168]

The solutions to the questions are

The probability that X is between 2 and 4 is 0.314
The probability that X exceeds 3 is 0.199
The expected value of X is 2
The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

$P(x) = \int\limits^a_b {f(x) \, dx$

So, we have:

$P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2$

Expand the expression

$P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}$

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

$P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3$

Expand the expression

$P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}$

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

$E(x) = \int\limits^a_b {x * f(x) \, dx$

So, we have:

$E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx$

This gives

$E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx$

Using an integral calculator, we have:

$E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}$

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

$E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx$

This gives

$E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx$

Using an integral calculator, we have:

$E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}$

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

brainly.com/question/15318348

#SPJ1

<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

7 0

2 years ago

The grocery store is having a candy sale. Originally, a candy bar was $1.50. The

dangina [55]

The markdown rate is 66.7%.

<h3>Markdown rate</h3>

Using this formula

Markdown rate= Change in price/original price×100

Let plug in the formula

Markdown rate=$1.50-$0.50/$1.50

Markdown rate=$1.00/$1.50×100

Markdown rate=66.66%

Markdown rate=66.7% (Approximately)

Inconclusion the markdown rate is 66.7%.

Learn more about markdown rate here:brainly.com/question/1153322

8 0

2 years ago