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4 years ago

Solve this system of linear equations. Separate the x- and y-values with a comma. -7x + 2y = -8 -16x + 9y = 26

1 answer:

8 0

(4,10).
1. There are three ways to solve this: elimination, substitution, graphing.
2. I chose elimination, so I had to get one negative variable and one positive variable of the same value (for example, 18 and -18)

-7x+2y=-8
-16x+9y=26
I chose to get 2y and 9y to equal -18y and 18y.
So, multiply the first equation by -9. Multiply the second by 2.
63x-18y=72
-32x+18y=52
the 18s cross each other out. So you're left with
63x=72
-32x=52. Add them.
31x=124, divide both sides by 31, and you'll get 4.
x=4
Plug your answer for x into one of the equations. Let's use the first one.
-7(4)+2y=-8
-28+2y=-8. add 28 to both sides.
2y=20, divide both sides by 2.
y=10.
This makes your answer (4,10)

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A change purse contains 16 coins. The coins are nickels, dimes, and quarters. The value of the change is $2.10. The number of di

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Taking into account the definition of a system of linear equations, there are 6 quarters in the change purse.

<h3>System of linear equations</h3>

A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.

Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.

<h3>Amount of quarters in the change purse</h3>

In this case, a system of linear equations must be proposed taking into account that:

N in the amount of nickels in the change purse.
D in the amount of dimes in the change purse.
Q in the amount of quarters in the change purse.

On the other hand, you know:

A change purse contains 16 coins.
The value of the change is $2.10.
The number of dimes is 4 fewer than the number of quarters.

The system of equations to be solved is

Equation 1: N + D + Q= 16

Equation 2: 0.05N + 0.10D + 0.25Q= 2.10

Equation 3: D= Q - 4

There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.

Substituting equation 3 into equation 1 you get:

N + Q - 4 + Q= 16 → N +2Q= 16 +4 → N + 2Q= 20

Isolating the variable N you get:

N= 20 - 2Q

Replacing this expression and equation 3 in equation 2 you get:

0.05×(20 -2Q) + 0.10×(Q -4) + 0.25Q= 2.10

Solving:

0.05×20 - 0.05×2Q + 0.10Q - 0.10×4 + 0.25Q= 2.10

1 - 0.10Q + 0.10Q - 0.40+ 0.25Q= 2.10

- 0.10Q + 0.10Q + 0.25Q= 2.10 -1 +0.40

0.25Q= 2.10 -1 +0.40

0.25Q= 1.50

Q= 1.50÷ 0.25

Finally, there are 6 quarters in the change purse.

Learn more about system of equations:

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5 0

2 years ago

A set of city temperatures in April are normally distributed with a mean of 19.7 degrees Celsius and a standard deviation of 2 d

Sati [7]

Answer:

19.77% of average city temperatures are higher than that of Cairo

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 19.7, \sigma = 2$