The cost to rent a moving van is $10 plus an additional $19 per hour. For $48, what is the

schepotkina [342]

1/4(x + 16) - x =

First, distribute the 1/4.

= 1/4x + 1/4 * 16 - x

= 1/4x + 4 - x

= 1/4x + 4 - 4/4x

= 1/4x - 4/4x + 4

= -3/4x + 4

$= -\dfrac{3}{4}x + 4$

8 0

3 years ago

HELP‼️‼️‼️‼️‼️‼️ 1. If 8x + 4y = 44, what is the value of 6x + 3y?

Gnom [1K]

Answer:

33.

Step-by-step explanation:

8x + 4y = 44

4(2x + y) = 44

2x + y = 44/4 = 11.

6x + 3y

= 3(2x + y)

= 3 * 11

= 33.

8 0

2 years ago

At the movie theatre, child admission is

Dmitriy789 [7]

Child = $5.70
Adult = $9.80

- We can conclude that the amount of adult is double the amount of child tickets bought. (total of $556.60)

For this, you will need to divide the total amount by two and draw a tree chart with each of the numbers to get the seperation of the two. I would find out how many 9.80s or 5.70s go into $56.60 so you'll just have $500

7 0

3 years ago

What is the answer to (-7)(2)(-2).

IgorC [24]

Answer:

28

Step-by-step explanation:

-2 × -7 = 14

14 × 2 = 28

thats the correct answer

8 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