Which properties did Marty use? Thank you very much!!

Solve the system of equations. - 9x + 4y = 6 9x + 5y = -33

liubo4ka [24]

Let us solve this system of equations by using the elimination method. Adding the 2 equations, we get

9x + 5y - 9x + 4y = -33 + 6

9y = -27

y = -3

Substituting this value of y in the first equation, we get

-9x + 4(-3) = 6

-9x - 12 = 6

9x + 12 = -6

9x = -18

x = -2

Therefore, x = -2, and y = -3. Hope this helps! If you have any questions, feel free to ask.

7 0

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.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(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

HELP!!! Name the space figure you can form from the net.

schepotkina [342]

I believe it is a triangular prism

3 0

2 years ago

The probability that John receives junk mail is 11 percent. If he receives 94 pieces of mail in a week, about how many of them c

SVETLANKA909090 [29]

Answer:

10 (10.34 rounded down)

Step-by-step explanation:

11% is the same as .11

Multiply by 94 for your answer.

.11 x 94 = 10.34

10 pieces

5 0

3 years ago

Given m||n, find the value of x. (4x-9)° (3x+29)°

anzhelika [568]

Answer:

38°

Step-by-step explanation:

4x-9=3x+29 ------- because corresponding angles are equal

x=38°

5 0

2 years ago