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2 years ago

What is the solution to the following system of equations?

Mathematics

1 answer:

Margaret [11]2 years ago

3 0

Hi there!

x + 2y = 2
x - 2y = -2

We need to solve x + 2y = 2 for y
x = 2y +2
Now substitute -2y + 2 for x in x - 2y =-2
x -2y = -2
-2y + 2 - 2y= -2
-4y + 2 = -2
-4y = -2 - 2
-4y = -4
y = -4/-4
y= 1

We gonna use the value of y to find x
We gonna find x by substitute 1 for y in x = -2y + 2
x = -2y + 2
x = -2(1)+2
x= -2 + 2
x = 0

Hence,
x = 0 and y = 1

So the correct option is (0,1)

I hope I helped!

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3 years ago

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Izzie recently took a math test where each correct answer (a) on the main part of the test was worth 8.3 points. She also correc

Temka [501]

Answer:

12 correct answers

Step-by-step explanation:

Since in the main part she scores 8.3 points for each question she answers correctly, we can assume that the number of questions she answers correctly=a

Therefor, the total number of points she achieved in the math test in the main part alone can be expressed as:

Total score(main part)=8.3×a=8.3a points

She also solved a bonus question worth=11 points

Consider expression 1 below

The total score in the whole test=Total score in the main part+Bonus points, where;

Total score in the whole test=110.6 points

Total score in the main part=8.3a points

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Substituting the values in expression 1:

8.3a+11=110.6

8.3a=110.6-11

8.3a/8.3=99.6/8.3

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3 years ago

Help what is 1/3 to decimal

sleet_krkn [62]

Hey there! I'm happy to help!

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3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$