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nevsk [136]
3 years ago
9

Express these numbers in standard form (a)5730(b)0.23(c)0.000231(d)7150000(e)500.8

Mathematics
1 answer:
alisha [4.7K]3 years ago
8 0

Answer:

5730 = 5.73×10³

0.23 = 2.3×10‐¹

0.000231 = 2.31×10-⁴

7150000 = 7.15×10⁶

500.8 = 5.008×10²

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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.
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We have the following distribution

x      0     1     2   3   4

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Part b

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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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