Identify the common factor in 6b

Given the following system of equations:

kompoz [17]

Answer:

(D) Divide the first equation, $-4x + 8y = 16$ , by 2.

Step-by-step explanation:

Given:

$-4x + 8y = 16 \ \ \ \ equation \ 1$

$2x + 4y = 8 \ \ \ \ equation \ 2$

We need to find the operation performed on equation so as to get resultant equation as:

$-2x + 4y = 8$

$2x + 4y = 8$

From Above we can see that there is no change in equation 2 with respect to resultant equation.

Also Resultant equation is simplified form of equation 1.

Simplifying equation 1 we get;

$-4x + 8y = 16$

We can see that 2 is the common multiple on both side.

Hence we will divide equation 1 with 2 we get

$\frac{-4x}{2}+\frac{8y}{2}=\frac{16}{2}\\\\-2x+4y=8$

which is the resultant equation.

Hence (D) Divide the first equation, $-4x + 8y = 16$ , by 2 is the correct option.

3 0

3 years ago

The difference of two number is 28. if one number is 11, find the other number

Leno4ka [110]

Answer:

39

Step-by-step explanation:

difference=28

other number=11

so:

11+28=39

39 is your other number

7 0

3 years ago

Read 2 more answers

Does someone know the answer to this √7x+16+4=x

dalvyx [7]

I think it is 54.332

8 0

3 years ago

Find the area of a regular decagon with a side length of 14 and an apothem of 25.

andriy [413]

Area = 10 * 1/2* * 14 * 25 (note decagon has 10 sides)

= 1750

4 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

3 years ago

Identify the common factor in 6b - 2b.