All categories

3 years ago

Draw the graph of the equation: 2(x–y)+3y=4

Mathematics

1 answer:

juin [17]3 years ago

4 0

Answer: 1

Step-by-step explanation:

see any combine like terms

2(x- y)+3y =4

2x - y+3y=4

subtract postive 3y with negative y

It gives you 2

now it 2x +2 = 4

-2

cancel the those 2

subtract 2 o

2x divide 2

gives you

x = 1

You might be interested in

I don’t know this one

torisob [31]

Answer:

3077.2

Step-by-step explanation:

A circle's mass is found by

Pie times R^2

7^2 is 49, times 3.14 is 153.86

then you Multiply 153.86 by 20 which gives you 3077.2

5 0

3 years ago

Read 2 more answers

I NEED HELP RNNNNNNNNNN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Law Incorporation [45]

The answer is the last option 32x8 hope this helps

3 0

2 years ago

Read 2 more answers

X/-3-41=-11 show work please

Brilliant_brown [7]

The correct answer is x = -90.

To find this, solve using the order of operations. See the example below.

x/-3 - 41 = -11 ----> Add 41 to both sides

x/-3 = 30 ----> Multiply each side by -3

x = -90

4 0

3 years ago

What is the independent and dependent variable

Orlov [11]

S is independent variable and T is the dependent variable

3 0

3 years ago

Read 2 more answers

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:

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

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

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

7 0

2 years ago