Write the equation of the line in slope-intercept form

Suppose you have a random variable that is uniformly distributed between 1 and 29. what is the expected value for this random va

Kryger [21]

Assuming the distribution is continuous, you have

$\mathbb E(X)=\displaystyle\int_1^{29}\frac x{29-1}\,\mathrm dx=\dfrac1{28}\int_1^{29}x\,\mathrm dx=15$

If instead the distribution is discrete, the value will depend on how the interval of number between 1 and 29 are chosen - are they integers? evenly spaced rationals? etc

3 0

3 years ago

The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability d

ollegr [7]

Answer: 0.70

Step-by-step explanation:

Given : The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:

x 0 1 2 3 4 5

P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05

Using the above probability distribution , the the probability that in a given week there will be at most 3 accidents is given by :_

$P(\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=0.20+0.30+0.20=0.70$

Hence, the required probability = 0.70

7 0

3 years ago

Bart used place value and the distributive property to rewrite the expression 6(412). His work is shown below. 6(412) = 6(400 +

egoroff_w [7]

Bart should;

Multiply 6(400), 6(10), and 6(2) ⇒ 2nd

Multiply each addend by 6 ⇒ 4th

Find the sum of (2400 + 60 + 12) ⇒ 5th

Step-by-step explanation:

Let us revise the distributive property

a(b + c) = ab + ac
a(b + c + d) = ab + ac + ad

Bart used place value and the distributive property to rewrite the

expression 6(412)

His work is shown 6(412) = 6(400 + 10 + 2)

We need to know what Bart should do to find the product

∵ 6(412) = 6(400 + 10 + 2)

- Multiply each number in the bracket by 6 in the right hand side

∴ 6(400 + 10 + 2) = (6)(400) + (6)(10) + 6(2)

∴ 6(400 + 10 + 2) = 2400 + 60 + 12

∴ He should to multiply each addend by 6 and then find the sum of

(2400 + 60 + 12)

Bart should;

Multiply 6(400), 6(10), and 6(2)

Multiply each addend by 6

Find the sum of (2400 + 60 + 12)

Learn more:

You can learn more about the place value in brainly.com/question/120752

#LearnwithBrainly

5 0

3 years ago

Read 2 more answers

A box contain 12 balls in which 4 are white 3 are blue and 5 are red.3 balls are drawn at random from the box.find the chance th

Anastasy [175]

Answer:

3/11

Step-by-step explanation:

From the above question, we have the following information

Total number of balls = 12

Number of white balls = 4

Number of blue balls = 3

Number of red balls = 5

We solve this question using combination formula

C(n, r) = nCr = n!/r!(n - r)!

We are told that 3 balls are drawn out at random.

The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!

= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 220 ways

The chance of selecting 3 balls at random = 220

To find the chance that all the three balls are selected,

= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls

Chance of selecting (white ball)= 4C1

Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1

Hence,

= [4C1 × 3C1 × 5C1]/ 220

= 60/220

= 6/22

= 3/11

The chance that all three are selected is = 3/11

4 0

3 years ago

What is the value of A when we rewrite<br> (5/2)^x +(5/2)^x+3 as A x (5/2)^x

elixir [45]

Answer:

Step-by-step explanation:

$(\frac{5}{2})^{x}+(\frac{5}{2})^{(x+3)}=(\frac{5}{2})^{x}+(\frac{5}{2})^{x}*(\frac{5}{2})^{3}\\\\=(\frac{5}{2})^{x}*[1+(\frac{5}{2})^{3}]$

Therefore,

$A=1+(\frac{5}{2})^{3}$

3 0

3 years ago