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Trava [24]
3 years ago
14

|14+x|-5=k solve for k​

Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

k=|14+x|-5

Step-by-step explanation:

just rewrite the equation as k=|14+x|-5

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How to tell the difference between multiplication and division word problems?
Sindrei [870]

Answer:

Multiplication is one of the four elementary mathematical operations of arithmetic, with the other ones being addition, subtraction and division. The result of a multiplication operation is called a product. And division is one of the four basic operations of arithmetic, the ways that numbers are combined to make new numbers. The other operations are addition, subtraction, and multiplication

5 0
2 years ago
Read 2 more answers
The number of people in a car that crosses a certain bridge is represented by the random variable X, which has a mean value <img
nexus9112 [7]

Answer:

E(Y) = 3 + 0.5 *2.7 = 3+1.35 = 4.35

Var (Y) = 0.5^2 Var (X) = 0.25 *(1.2) = 0.3

And then we have the following parameters for the random variable Y

\mu_Y = 4.35 , \sigma^2_Y = 0.3

And the best option is:e. mean = $4.35, variance = $ .30.

Step-by-step explanation:

For this case we know the following info " The toll on the bridge is $3.00 per car plus $ .50 per person in the car"

So then we can create the following linear function:

Y = 3 + 0.5 X

Where Y represent the total amount of money that is collected from a car. And X represent the number of people in the car.

We have the parameters for the random variable X given by:

E(X) = \mu_X = 2.7 , \sigma^2_x = 1.2

So then we can find the expected value for the random variable Y like this:

E(Y) = 3 + 0.5 E(X)

And if we replace the expected value for X we got this:

E(Y) = 3 + 0.5 *2.7 = 3+1.35 = 4.35

Now we can find the variance for the random variable Y like this:

Var (Y) = Var (3) + 0.5^2 Var(X) + 2 Cov(3, 0.5X)

We know that for a number we don't have variance since is a constant and the covariance between a number and a random variable is 0 so then we have just this for the variance of Y:

Var (Y) = 0.5^2 Var (X) = 0.25 *(1.2) = 0.3

And then we have the following parameters for the random variable Y

\mu_Y = 4.35 , \sigma^2_Y = 0.3

And the best option is:e. mean = $4.35, variance = $ .30.

4 0
3 years ago
I'm marking answers as brainliest ----- One of the vertices for this linear programming problem is __________. ----- (1,2) (3.5,
Travka [436]

Answer:

(0,2.3)    if only one choice expected.

(0,2.3) and (0,0)  if "all that apply" expected, since (0,0) is also a vertex of the green region.

Step-by-step explanation:

From the attached diagram, the given answer optsions are shown in brown.

The red line shows the optimal combinations of the two variables, hence a maximization problem.

Out of the four answer options, two coincide with a vertice of the valid region (shown in green), namely (0,0) and (0, 2.3).

Out of the two indicated options which correspond to a vertex, the origin (0,0) is usually ignored for a maximization problem, which leaves us with (0,2.3) as the answer.

3 0
3 years ago
A bowler knocks at least 6 pins 70% of the time.Out of 200 rolls, how many times can you predict the bowler will knock down at l
ankoles [38]

The bowler will hit 6 pins 140 times because 70% of 200 is 140

6 0
3 years ago
Jane had of a meter of ribbon. She used of a meter of ribbon to decorate a card. How much ribbon did she have left after she dec
emmainna [20.7K]
She had none left. if she only had a meter and then used a meter there would be none left.
5 0
2 years ago
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