What is the y-intercept of the function f(x) =

7. Susie bought some reams of paper for 5 each and a $200 printer. He spent a minimal of $450. write and solve an equation to fi

Dovator [93]

Answer:

$450-200÷5=the number of reams of paper Susie purchased.

Step-by-step explanation:

$450-$200=$250

$250÷5= 50

Therefore, Susie purchased 50 reams of paper

3 0

2 years ago

Ernesto Montenegro purchase a twelve 1000 Bonds having a quoted price of 100.75 he have to pay a 6% brokerage fee of the same pr

LenKa [72]

Answer: $106.8

Step-by-step explanation:

Given that the bond price is $100.75

With a 6% brokerage fee

Find the total cost to the nearest cent

First, we calculate the 6% brokerage fee= 6/100 × 100.75

= 0.06 x 100.75

= 6.045

Total cost therefore is the sum of brokerage fee and the cost of bond

= $6.045 + $100.75

= $106.795

To the nearest cent = $106.8

5 0

3 years ago

A side of a square is 8 3/2 inches. Using the area formula A = s2, determine the area of the square.

Sedaia [141]

We are given that:
side length = 8 3/2 inches = 9.5 inches
the rule for the area is: A = s^2

Substitute with the side length in the rule given to get the area as follows:
Area = s^2 = (9.5)^2 = 90.25 inch^2

4 0

2 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$