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3 years ago

Which equation can pair with 3x + 4y = 8 to create a consistent and independent system?

1 answer:

6 0

There isn't any choices given in this problem but I can give you a tip to answer these kinds of problem. When you say independent, it means that the equation cannot be algebraically altered to result to 3x + 4y = 8. So the equation 6x + 8y = 16 is not an independent equation since when you divide each side by 2 it will result to 3x + 4y =8. You get the idea. <span />

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Which expression is equivalent to the expression shown?

Basile [38]

Answer:

5^0

Step-by-step explanation:

(5^3)^9 = 3 + 9 = 12

= 5^12

5^12/5^12 = 12 - 12 =0

5^0

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3 years ago

Write an equation in point-slope form for the line that passes through the point with the given slope. point: (4,2) and m=3

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Answer:

y = 3x - 10

Step-by-step explanation:

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3 years ago

What is 9.27 million in scientific notation?

Alex787 [66]

9,270,000 = 9.27 * 10^6

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3 years ago

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732,178+167-542,137=

madreJ [45]

( 732,178 + 167 ) = 899,178
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3 years ago

Read 2 more answers

An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations

Hoochie [10]

Answer:

a) $E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

b) $E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

Part a

We have the following distribution function:

Y 0 1 2 3

P(Y) 0.45 0.2 0.3 0.05

And we can calculate the expected value with the following formula:

$E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

Part b

For this case the new expected value would be given by:

$E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)$

And replacing we got

$E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

5 0

3 years ago