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MA_775_DIABLO [31]
3 years ago
11

A drawer contain pens , 14 blue pens and 8 black pens , 14 silver paper clips , and 16 white paper clips . If you randomly selec

t a pen and paper clip from the drawer , what is the probability that you select a red pen and white paper clip
Mathematics
1 answer:
vichka [17]3 years ago
4 0

Step-by-step explanation:

Total pens = 22

Total clips = 30

Probability of red pen = 0

Because there are no red pens

Probability of white clip = 16/30 = 8/15

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Vitek1552 [10]
To solve this we need only substitute values of u and w, and then calculate.

<span>u+5v-w = 6 + 5v - 9 = 5v +6 - 9 = 5v - 3</span>
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Which sum will be neither positive nor negative. a. 3+2 b. -4+(-4) c. -6+6 d. 7+(-6)​
anastassius [24]

Answer:

c

Step-by-step explanation:

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3 years ago
Write the following as a fraction in simple form <br><br>1.6​
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Hi!
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2 years ago
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

so

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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I need help ASAP! It's urgent.. PLISSSSS ​
Soloha48 [4]

Answer:

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