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2 years ago

A) Rearrange the equation to make w the subject. 2(1 + W) = P

Mathematics

1 answer:

insens350 [35]2 years ago

4 0

Answer:

Step-by-step explanation:

$2(1 + W) = P\\Divide -both-sides-of-the-equation-by ; 2\\\frac{ 2(1 + W) }{2} = \frac{P}{2} \\1+ W = P/2\\W = \frac{P}{2} -1\\\\(b .) 2y = 10x + 6\\Re -write -in -slope-intercept-form : y= mx+b\\\frac{2y}{2} =\frac{10x}{2} +\frac{6}{2} \\y = 5x + 3\\Gradient = 5$

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tarik starts a morning walk at 7:00 am. his distance, in miles, to the nearby grocery store after x hours is f(x)

Grace [21]

Total distance walked by Tarik is 0.75 miles according to the given equation.

What is equation?

In arithmetic, an equation could be a formula that expresses the equality of 2 expressions, by connecting them with the sign =.

Main body:

Given that

The expression that represents the Tarik's distance is

=> f(x) = 3∣x−0.5∣

And also given that at 7.15 am he is some distance away from the store

Let Tarik is 'a' miles away from the store at 7.15 am

The time taken to walk 'a' miles = 7.15 - 7.00 = 15 minutes

As we know 60 minutes = 1 hr

=> 15 minutes = 15/60 = 0.25 hrs

The given expression to find the Distance is

f(x) = 3 ∣x − 0.5∣ where x is number of hours

At x = 0.25

f(0.25) = 3 |0.25 - 0.5|

= 3 |- 0.25|

= 0.75 miles

Therefore,Tarik is 0.75 miles away from the store.

To know more about equation , visit:

brainly.com/question/22688504

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8 0

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