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3 years ago

What is the value of the expression below when x 3 and y = 8? 6x + 10y

Mathematics

2 answers:

suter [353]3 years ago

7 0

Answer:

Step-by-step explanation:

So we have the expression:

$6x+10y$

And we want to evaluate it when x is 3 and y is 8.

So, substitute:

$=6(3)+10(8)$

Multiply:

$=18+80$

Add:

$=98$

And we're done!

schepotkina [342]3 years ago

5 0

6(3) + 10(8)

18 + 80

=

98

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Are the ratios 25/45 and 15/27 proportional? Explain.

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3 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

kobusy [5.1K]

Answer:

The probability the die chosen was green is 0.9

Step-by-step explanation:

From the information given :

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.

SO, 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.

When 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:

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

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

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

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

$=\dfrac{12}{2 \times 1}\times (\dfrac{1}{6})^4$

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

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

$= \dfrac{1}{216}$

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

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

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

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

$= 6 \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= \dfrac{9}{216}$

∴ 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 = $(\dfrac{1}{216} \times \dfrac{1}{2} )+ ( \dfrac{9}{216} \times \dfrac{1}{2})$

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

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

Using Bayes Theorem; the probability that the die was green can be computed as follows:

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)

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 ) = $\dfrac{\dfrac{1}{48} }{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{1}{48} \times \dfrac{216}{5 }$

P(second die (green) | two 1's and two 4's ) = $\dfrac{9}{10}$

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

∴

The probability the die chosen was green is 0.9

3 0

3 years ago