I need help finding the answer for this.

I need help!! thnaks

liraira [26]

Answer:

7a + 2b

Step-by-step explanation:

Given

4a + 3b + 3a - b ← collect like terms

4a + 3a + 3b - b

= 7a + 2b

3 0

2 years ago

the area of a rectangle is 95 square yard. if the perimeter is 48 yards find the length and width of rectangle

Vladimir [108]

Try this option:
according to the condition w+l=48 and w*l=95.
Using these two equation:
$\left \{ {{l+w=48} \atop {lw=95}} \right. \ =\ \textgreater \ \ \left[\begin{array}{ccc}l=24- \sqrt{481};w=24+ \sqrt{481}\\ \\l=24+ \sqrt{481};w=24- \sqrt{481}\end{array}$

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 ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

A mechanic gives a survey to all of his customers asking them to rate the quality of the service they received. He then keeps tr

topjm [15]

Answer:

Some customers who rate their service as high quality return for additional services.

Step-by-step explanation:

Last year, the results showed that of the customers who reported high quality service, 20% returned for additional services.

This shows than some customers are satisfied overall and return back for additional services.

6 0

2 years ago

Samantha’s college runs on a trimester schedule so she receives a bill 3 times a year for tuition. Each trimester costs $1,450,

SOVA2 [1]

Answer:

The total cost of Samantha's college degree is approximately $10,800.

Step-by-step explanation:

Cost of each trimester = $1,450

Now, total cost of fees per year = 3 x Cost of each trimester

= 3 x $1,450 = $4350

So, cost of fees in 2 complete years = 2 x $4,350 = $8,700

Also, Cost of books each trimester =$350

Now, total cost of books per year = 3 x Cost of each trimester

= 3 x $350 = $1,050

So, cost of books in 2 complete years = 2 x $1,050 = $2,100

So, total cost of Samantha's college = Total Trimester fees cost +Total books cost

= $8,700 + $2,100 = $10,800

Hence, the total cost of Samantha's college is approximately $10,800.

3 0

3 years ago

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I need help finding the answer for this.​

I need help finding the answer for this.