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2 years ago

WILL MARK BRAINLIEST TO THE FIRST PERSON

Mathematics

2 answers:

WINSTONCH [101]2 years ago

6 0

Answer:

Step-by-step explanation:

7^2 = (2(6)^1/2)^2 + BC^2 (Pyth. theorem)

BC = 5

maxonik [38]2 years ago

5 0

Answer:

its supposed to be 6.08 but we can take of..08 so 6

Step-by-step explanation:

$a + b = c \\ ( 2 \sqrt{6} ) {}^{2} + {b}^{2} = {7}^{2} \\ 2 \times 6 + {b}^{2} = 49 \\ {b}^{2} = 49 - 12 \\ {b }^{2} = 37 \\ \sqrt{b} {}^{2} = \sqrt{37} \\ b = 6.08 \: or \: 6$

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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:

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

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Ede4ka [16]

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Answer:

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Answer:

Step-by-step explanation:

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hoa [83]

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2 years ago

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