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3 years ago

3. A bag has five blue marbles, seven red marbles and nine green marbles. Three marbles are drawn

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

7 0

Answer:

The probability is 9/266

Step-by-step explanation:

The first thing to do here is get the total number of marbles = 5 + 7 + 9 = 21 marbles

Now we are drawing the marbles without replacement, this means once a marble is drawn there is a decrease in the total number of marbles left.

The probability of drawing a green marble initially = number of green marbles/ total

number of marbles = 9/21 = 3/7

Now, we want to draw a blue marble

There are 5 blue marbles and we have already drawn one marble from the total. So the probability of drawing a blue marble would be 5/20 = 1/4

And lastly, another green marble

That would make 6 green marbles left and a total of just 19 left. So the probability at this point for green marbles = 6/19

Having and in probability means we multiply all

So this means the probability we are calculating is; 3/7 * 1/4 * 6/19 = 18/(28 * 19) = 9/(14 * 19) = 9/266

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Read 2 more answers

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

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tresset_1 [31]

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