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sveticcg [70]
3 years ago
10

3 for performance, 5 for voice, 5 for participation, and 11 for dance. She randomly picks a trophy for show and tell day at scho

ol. What is the probability it will be for dance?
Mathematics
1 answer:
joja [24]3 years ago
4 0
Add all together and then put dance as numerator, 11/24
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A sedan will drive ten miles north, 15 miles east, 13 miles south, 15 miles west, and 22 miles
pishuonlain [190]
The total distance is 75 miles
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3 years ago
Solve for A <br> ASAP ANSWER
vlada-n [284]

Step-by-step explanation:

A = B

7x + 40° = 3x + 112°

7x - 3x = 112° - 40°

4x = 72°

x = 18°

A = 7x + 40°

= 7(18°) + 40°

= 126° + 40°

= 166°

hope it help

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5 0
2 years ago
What is p ÷ 6 = 9 ? I'm not really understanding it
Shalnov [3]
Let's figure this out. Step by step:

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8 0
3 years ago
Read 2 more answers
Keith as 11/12 hours to play, he has already played 1/4. which is the best estimate of the fractional part of an hour he has lef
____ [38]

Answer:

The best estimate is greater than 1/2 but less than 3/4

Step-by-step explanation:

If you multiply 1/4 by 3/3 (which is also 1), you can change the denominator without changing the value.  So 1/4 is equal to 3/12.

Since keith has 11/12 hours to play, and he has already played 3/12 hours, subtract 3/12 from 11/12 to get 8/12 hours. This is how much time he has left to play.  

If you simplify 8/12 hours, you get 2/3 hours.

So the best estimate would be: greater than 1/2 but less than 3/4.

4 0
3 years ago
At what angle does a diffraction grating produce a second-order maximum for light having a first-order maximum at 20.0 degrees?
hichkok12 [17]

Answer:

At 43.2°.

Step-by-step explanation:

To find the angle we need to use the following equation:

d*sin(\theta) = m\lambda

Where:

d: is the separation of the grating

m: is the order of the maximum

λ: is the wavelength

θ: is the angle              

At the first-order maximum (m=1) at 20.0 degrees we have:

\frac{\lambda}{d} = \frac{sin(\theta)}{m} = \frac{sin(20.0)}{1} = 0.342

Now, to produce a second-order maximum (m=2) the angle must be:

sin(\theta) = \frac{\lambda}{d}*m

\theta = arcsin(\frac{\lambda}{d}*m) = arcsin(0.342*2) = 43.2 ^{\circ}

Therefore, the diffraction grating will produce a second-order maximum for the light at 43.2°.    

I hope it helps you!                                                        

6 0
3 years ago
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