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4 years ago

P=2L+2w solve for w helppppp

Mathematics

2 answers:

Gala2k [10]4 years ago

6 0

P=2l+2w

Subtract 2l from both sides
P-2l=2w

Divide all terms by 2
(1/2)P-l= w

Final answer: w= (1/2)P-l

AnnZ [28]4 years ago

4 0

L=-w+p/2 i found it on auto calculator

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The correct answer is A.

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3 years ago

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77julia77 [94]

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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) $P(X=0) = 0.2$

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

And replacing we got:

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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

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andreyandreev [35.5K]

Answer:

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∠ AFE = 80°

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Darya [45]

Answer:

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Step-by-step explanation:

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3 years ago

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