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Sorry, my bad... but it still says 14 is incorrect.

What do you mean 38,416^(1/4) is NOT 14?. Enter 38,416 in a calculator and take its square root TWICE, you should get 14.

It says this is incorrect, plus 38416^1/4 is not 14. Does anyone have an answer that they know is right and that they can explain?

Since 98 =2 x 7^2, it therefore follows that if we multiply 98 x [2^3 x 7^2] =98 x 392 =

[98 x 392]^1/4 =[ 2^4 x 7^4]^1/4 =2 x 7 = 14.

I'm not sure how to do this, but from your definition what continuous is, I could attempt it. 

Set the two functions equal to each other to find where they intersect, so you can draw them w/o lifting your pencil once you reach that point(s).

5x²+2=11x
5x²-11x+2=0

x=2, x=1/5
 

The smallest value of x where they intersect is 1/5, so a=1/5.

Graphed it and it seems to work so probably a=1/5.

You basically have it, all you have to do is set the denominator equal to 0 and isolate the x. 
3x+2=0 --> x=-2/3
x+1=0 --> x=-1
There's your a and b.
Add them.
-1+(-2/3)=-5/3

a) 
Distance is 195 miles.
Speed or rate is 50 mph

1) Use the distance formula.
d=rt; where d=distance, r=rate or speed, t=time
You have d and r, solve for t.
195=50t
t=195/50=3.9 hours
 

2) Tack on that time to the time it leaves. 

Leaves at 3:30

Add the whole amount of hours (3)
Becomes 6:30
Turn the 0.9 hours to minutes by multiplying by 60
0.9*60=54 minutes
Add 54 minutes to 6:30, you get the arrival time which is:
7:24

b) Since the coach has a lower average speed in other words, it's slower, the arrival time will be later than the calculated time of 7:24.

You’re right; I did misunderstand it. Moreover, I jumped to criticism to soon.  The reason I gave for his incorrect solution is wrong. His solution does imply a probability for a ball going into an already occupied cup. 

There is still a problem though.

In my initial analysis, I calculated the probabilities based on indistinguishable balls.

Carl Bryan’s solution method assumes distinguishable balls.  These are two different probabilities.

(Note that the cups are always distinguishable, else there is only one-way to distribute the balls for a success.)

Carl Bryan’s solution (where both the balls and cups are distinguishable) in binomial form is

\(\dfrac {C!}{(C-B)!} \div (C^B)\\ \dfrac {10!}{(7!)} \div (10^3) = \frac{18}{25} = 0.72 \)

My method is

The number of ways to select three (3) cups from a set of 10 (one for each ball) is 10C3 = 120

The number of ways to select two (2) cups from a set of 10 (one for two balls and one for a single ball) is 10C2 =45

The number of ways to select two (2) cups from a set of 10 (one for single a ball and one for two balls) is 10C2 =45

(Note the selection of two cups is counted twice because the cup with the single ball can be switched with the pair in the other cup)

The number of ways to select one (1) cup from a set of 10 (one for all three balls) is 10C1 = 10

Total number of distributions = 220

Number of successes = 120

120/220 = (6/11) ≈  0.5455

In pure binomial form it is:

\(\dfrac {C!}{(C-B)!(B!)} \div \dfrac {(C+B-1)!}{( C-1)!(B!)}\\ \dfrac {10!}{(7!)(3!)} \div \dfrac {12!}{(9!)(3!)} = \dfrac{6}{11}\)

Now the question: what is the correct way to interpret the balls? Should they be indistinguishable or distinguishable?  Is it mathematician’s choice?

I choose indistinguishable because Ping-pong balls are indistinguishable and there is nothing in the question to indicate they are distinguishable.

Carl Bryan’s solution method is for distinguishable balls, he may not have considered the indistinguishable.

What is the default? Or is there one?

Comments ................Alan, CPhill, Melody and/or anyone who is versed on these questions and their solutions.

GA

Probability of taking a red sweet = 3/8
eats it so amount of reds in the bag drops to 2, total amount of sweets in bag drops to 7.
Probability of taking the 2nd red sweet = 2/7

P(R,R) = 3/8(2/7) = 6/56

Do you mean: abs[-5^2 - 16] =9? If so, why is W/A giving 41 as the answer? This is how they come to that answer:

Simplify the following:
abs(-5^2 - 16)

Evaluate 5^2.
5^2 = 25:
abs(-25 - 16)

Evaluate -25 - 16.
-25 - 16 = -41:
abs(-41)

Absolute value gives the distance to the origin.
Since -41<=0, then abs(-41) = 41:
=41

https://www.wolframalpha.com/input/?i=abs%5B-5%5E2+-+16%5D+%3D%3F

No, there is only a finite number of those numbers.

proof: suppose |x²-16| is a prime number. |x²​-16|=|x-4|*|x+4|, so one of the multiplied numbers=1 (because the product is a prime number).

|x-4|=1 means x=5 or x=3, and |x+4|=1 means x=-5 or x=-3. if x=5 then |x+4|=9, and because 9 is not a prime number (9=3*3) x cannot be equal to 5.

if x=-5 then |x-4|=9, so x cannot be equal to -5.

so if |x²-16| is a prime number, then x=3 or x=-3. the product of the solutions is 3*-3=-9, and that's the answer.

You should be ashamed for your incorrect answer, GUEST!

\(\frac{5}{24}+|x-\frac{11}{48}|<\frac{5}{16}\)

1) Isolate the absolute value part
\(|x-\frac{11}{48}|<\frac{5}{48}\)

2) For absolute value inequalities, split the inequality into two parts, then solve.

For one of the parts, just drop the absolute value and solve.
For the second part, drop the absolute value, flip the inequality sign and make the other side negative. 
\(x-\frac{11}{48}<\frac{5}{48}\\ x<\frac{16}{48}\\ x<\frac{1}{3}\) and  \(x-\frac{11}{48}>-\frac{5}{48}\\ x>\frac{6}{48}\\ x>\frac{1}{8}\)
3) Turn that into interval notation.

(1/8)

\((\frac{1}{8},\frac{1}{3})\)

8) You have to give total number of horses and emus in order to get a unique answer, otherwise, there could be many solutions to the problem.

9) Let the number of Fancy skirts = F

Then the number of plain skirts = 7 - F

28F + 15[ 7 - F] = 131

28F + 105 - 15F = 131

13F = 131 - 105

13F = 26

F = 26 / 13

F = 2 -Fancy skirts.

7 - 2 = 5 - Plain skirts.

Well, if they are not going to include Fred, then they only have 9 people to choose from.

So, 9 choose 3 =9C3 = 9! / [(9 - 3)! . 3!] =84 possible committees.

((8 + 2)!)/(8! 2! 2^(8 + 2)) = 45/1024 ≈ 0.04395 =4.395% probability of exactly 8 heads.

A.

(10 – 4) + 2

= 6 +2 

= 8

B.

10 – 4 + 2

= 10 + 2 - 4

= 12-4

= 8

C.

10 – (4 + 2)

= 10 - 6

= 4

D.

(10 – 4 + 2)

=> same as B

So, C is the correct answer

The two numbers can be +3 and -5. Also -9 and -11 will work. There maybe an infinite number of them.

abs[3^2 - 16] = 7 which is a prime number.

abs[-5^2 - 16] =41 which is a prime number.

+3 x -5 =-15

C -  10 - (4 + 2)  do the brackets first.

10 - (6) = 4

This is my understanding of this problem:

There are the following consecutive possibilities:

123, 234, 345, 456, 567, 678, 789, 8910. Or 8 possibilities.

The probability of any of the above combination is:

1 x 1/9 x 1/8 =1/72

1/72 x 8 =8/72 = 1/9 probability of drawing 3 consecutive numbers. Therefore:

1 - 1/9 =8/9 probability of drawing 3 non-consecutive numbers.

P.S. Somebody should check this!.

Your final total should be: 35/126 =5/18

You need to re-read carl bryan's answer- because you misunderstood it

4)
Find AC.
Picture: https://latex.artofproblemsolving.com/9/1/d/91d520533f0fb8cb325932013b2aebf0db9bde54.png

cos-rule

\(\begin{array}{|rcll|} \hline \mathbf{(7\sqrt{3})^2 }& \mathbf{=}& \mathbf{ 9^2 + AC^2 -2\cdot 9 \cdot AC \cdot \cos(150^{\circ}) } \\ 49\cdot 3 &=& 81 + AC^2 - 18 \cdot AC \cdot \cos(150^{\circ}) \quad & | \quad \small{\cos(150^{\circ}) = -\dfrac{\sqrt{3}}{2}} \\ 49\cdot 3 &=& 81 + AC^2 + 18 \cdot AC \cdot \dfrac{\sqrt{3}}{2} \\ 147 &=& 81 + AC^2 + 9 \cdot \sqrt{3} \cdot AC \quad & | \quad - 81 \\ 66 &=& AC^2 + 9 \sqrt{3} \cdot AC \\ AC^2 + 9 \sqrt{3} \cdot AC - 66 &=& 0 \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{ (9\sqrt{3})^2 - 4\cdot (-66) } } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{ (81\cdot 3 + 264 } } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{507} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{13^2\cdot 3} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} + 13\sqrt{3} } {2} \quad & | \quad \text{AC}> 0\ !\\\\ AC &=& \dfrac{4\sqrt{3} } {2} \\\\ \mathbf{AC} & \mathbf{=} & \mathbf{ 2\sqrt{3} } \\ \hline \end{array}\)

This is the the wrong blòódy solution for this question! (MAYBE?)