Order the values from least to greatest pi,√11,3.4,3.1

VLD [36.1K]

The answer is 2 two(2x2)=4

6 0

3 years ago

Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.

Sindrei [870]

Answer:

a) $P(X=3) = 0.1$

b) $P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

c) $P(X=4) = 0.3$

d) $P(X=0) = 0.2$

e) $E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

f) $E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

Step-by-step explanation:

We have the following distribution

x 0 1 2 3 4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

$P(X=3) = 0.1$

Part b

We want this probability:

$P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

Part c

For this case we want this probability:

$P(X=4) = 0.3$

Part d

$P(X=0) = 0.2$

Part e

We can find the mean with this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

Part f

We can find the second moment with this formula

$E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

4 0

3 years ago

X(x-y) - y( x- y) simplify

Rudik [331]

Step-by-step explanation:

x²-xy-xy+y²

x²+2xy+y²

hope it helps

3 0

3 years ago

Miriam is picking out some movies to rent, and she is primarily interested in comedies and foreign films. She has narrowed down

Keith_Richards [23]

Answer:

There are 795 combinations.

Step-by-step explanation:

The number of ways or combinations in which we can select k element from a group of n elements is given by:

$nCk=\frac{n!}{k!(n-k)!}$

So, if Miriam want to choose 3 movies with at least two comedies, she have two options: Choose 2 comedies and 1 foreign film or choose 3 comedies.

Then, the number of combinations for every case are:

1. Choose 2 Comedies from the 10 and choose 1 foreign film from 15. This is calculated as:

$10C2*15C1=\frac{10!}{2!(10-8)!}*\frac{15}{1!(15-14)!}$

$10C2*15C1=675$

2. Choose 3 Comedies from the 10. This is calculated as:

$10C3=\frac{10!}{3!(10-3)!}=120$

Therefore, there are 795 combinations and it is calculated as:

675 + 120 = 795

8 0

3 years ago

A. [4 marks]

storchak [24]

The value of d is -2

Step-by-step explanation:

The nth term of the arithmetic sequence is $u_{n}=u_{1}+(n-1)d$ , where

$u_{1}$ is the first term
d is the common difference between each 2 consecutive terms

The nth term of the geometric sequence is $u_{n}=u_{1}r^{n-1}$ , where

$u_{1}$ is the first term
r is the common ratio between each two consecutive terms $r=\frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}$

∵ $u_{1}$ of an arithmetic sequence is 1

∵ The common difference is d, where d ≠ 0

∵ $u_{2}$ , $u_{3}$ , $u_{6}$ are the first 3 terms of a geometric sequence

∵ $u_{2}$ = 1 + (2 - 1)d

∴ $u_{2}$ = 1 + d

∵ $u_{3}$ = 1 + (3 - 1)d

∴ $u_{2}$ = 1 + 2d

∵ $u_{6}$ = 1 + (6 - 1)d

∴ $u_{2}$ = 1 + 5d

∴ The first 3 terms of the geometric sequence are (1 + d) , (1 + 2d) ,

(1 + 5d)

∵ The common ratio in the geometric sequence is $r=\frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}$

∴ $r=\frac{1+2d}{1+d}=\frac{1+5d}{1+2d}$

- Use cross multiplication with the equal fractions

∵ $\frac{1+2d}{1+d}=\frac{1+5d}{1+2d}$

∴ (1 + 2d)(1 + 2d) = (1 + d)(1 + 5d)

∴ 1 + 2d + 2d + 4d² = 1 + 5d + d + 5d²

- Add like terms in each side

∴ 1 + 4d + 4d² = 1 + 6d + 5d²

- Subtract 1 from both sides

∴ 4d + 4d² = 6d + 5d²

- Subtract 4d from both sides

∴ 4d² = 2d + 5d²

- Subtract 4d² from each side

∴ 0 = 2d + d²

- Take d as a common factor

∴ 0 = d(2 + d)

- Equate each factor by 0

∴ d = 0 but we will reject it because d ≠ 0

∴ 2 + d = 0

- Subtract 2 from both sides

∴ d = -2

The value of d is -2

Learn more:

You can learn more about the sequence in brainly.com/question/1522572

#LearnwithBrainly

7 0

3 years ago