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olga55 [171]
3 years ago
10

Solve: X + y = 4 -5x - 6y = -18

Mathematics
2 answers:
OLga [1]3 years ago
7 0

Answer:

x = 6 and y = -2

Step-by-step explanation:

In an equation our aim is to find the value of what we are looking for as well as keeping the equation balanced. For example if we took away 5 only from one side then the equation would change so it's an important rule to keep in mind when solving equations, that you need to keep both sides of the equation the same.

x + y = 4

-5x - 6y = -18

→ We have to multiply the top equation by -6 so the y coefficients are the same

-6x - 6y = -24

-5x - 6y = -18

→ Minus the equations

-x = -6

x = 6

→ Substitute x = 6 back into x + y = 4

6 + y = 4

y = -2

So the solutions are y = -2 and x = 6

LiRa [457]3 years ago
4 0

Answer: x = -5, y = 9

Step-by-step explanation:

The solution to the system of equations can be represented as a point.

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vampirchik [111]

Answer:

The minimum score required for an A grade is 90.

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 problem, we have that:

\mu = 78.9, \sigma = 9.3

Find the minimum score required for an A grade.

Top 12%, so at least the 100-12 = 88th percentile, which is the value of X when Z has a pvalue of 0.88. So it is X when Z = 1.175.

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

1.175 = \frac{X - 78.9}{9.3}

X - 78.9 = 1.175*9.3

X = 89.8

Rounding to the nearest whole number, the answer is 90.

The minimum score required for an A grade is 90.

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