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Harman [31]
2 years ago
9

Anybody have an answer to this ??

Mathematics
1 answer:
kondaur [170]2 years ago
7 0

Answer:

c

Step-by-step explanation:

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One side of a square is 5n inches long. Another side is n + 16 inches long. How long is a side of the square?
LiRa [457]
Its a square.
So 5n = n + 16
Therefore 4n = 16 (taking one n from each side).
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Double check by putting 4 in the equation where n is, and it makes sense, so that is the right answer.
Your answer is c.
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Write the equation in function form. Show your steps. <br> 2x-3y=0
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Answer: Heyaa!

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Step-by-step explanation:

Substitute the value of the variable into the equation and simplify.

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2 years ago
If you received a 5% raise and your final paycheck was 100(which includes your 5% raise), how much was your original paycheck?
Schach [20]

so you were earning say "x", so "x" is the 100% of  your paycheck.

but you're da bomb and so you got a raise of 5%, so the new amount of your paycheck is 100% + 5%, so is 105% really, and we happen to know that is $100.


\bf \begin{array}{ccll} \$&\%\\ \cline{1-2} 100&105\\ x&100 \end{array}\implies \cfrac{100}{x}=\cfrac{105}{100}\implies 10000=105x \\\\\\ \cfrac{10000}{105}=x\implies 95.238\approx x

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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

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
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