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i have figured it out

There are 6 multiples of 8 below 50. 8, 16, 24, 32, 40, 48. There is a 44/50 chance that he doesn't choose 8 the first time. The second time is also a 44/50 chance he doesn't choose 8. so the probability that he doesn't choose at least 1 multiple of 8 is (44/50)^2=0.7744. so the probability that he chooses at least 1 multiple of 8 is 1-0.7744=\(\boxed{0.2256}\)

https://web2.0calc.com/questions/help_54234#r3

I googled it, and apparently, there's an operator called used in coding that would work. It's called operator \(\text{%}\), and it finds the remainder when division is performed. 

It would work here, because \(3 \times 4 = 12\), and the remainder when 12 is divided by 5 is 2.

Thus, the missing operator is \(\color{brown}\boxed{\text{%}}\)

Here is a link to learn more: https://www.sciencedirect.com/topics/engineering/arithmetic-operator

That's smart actually! I'll do that next time!!

Choose  any of the odd integers  for the units place   =    3C1 =  3

Then we can choose any of the remaining four and permute them in 2 ways = 4P2   =  12 

So

3 * 12  =     36   three-digit integers

Hi Rudi,

A number is not a missing operator.

I think it must be some new operator, that is rarely used.  One I have not heard of. 

Here is another possibility but still more than one operator has been introduced.

\(\lfloor 3*4\div 5 \rfloor=2\)

0.18B  +  0.35*[85 - B] ==0.19 * 85, solve for B

B ==80 Liters of 18% brine

85 - 80 ==5 Liters of 35% brine.

Complete the expression by filling in the missing operator.

Hello Guest!

\(3 * 4 [BLANK] 5 =2-->3 * 4\ \ {\color{OliveGreen}-\ 2*}\ 5\ =2\)

  !

Repeated here.

Hi Builderboi.

Please answer repeat questions on the original thread.

Then put a not on the repeat that you have done so.

You can give yourself a point for both answer posts.

Because \(\triangle ABC\) is isoceles, \(AB = 18 \sqrt 2 \)

Because it is an isceles right triangle, \(\angle B = 45\), meaning that \(BY = {12 \over \sqrt2} = 6 \sqrt 2 \)

We also know that \(AY = 12 \sqrt 2\), because \(BY + YA = AB\)

Thus, the ratio of \(BY: YA = 6 \sqrt 2: 12 \sqrt 2 = \color{brown}\boxed{1 \over 2} \)

There are  6 x 6 x 6  =  216  ways for the dice to fall.

The ways to get a sum of 10:

6,3,1     5,4,1     4,5,1     3,6,1     2,6,2     1,6,3

6,2,2     5,3,2     4,4,2     3,5,2     2,5,3     1,5,4

6,1,3     5,2,3     4,3,3     3,4,3     2,4,4     1,4,5

             5,1,4     4,2,4     3,3,4     2,3,5     1,3,6

                          4,1,5     3,2,5     2,2,6

                                       3,1,6

The probability is  27/216

Simplify the equation to: \(4x^2-4x+1=0\)

We can rewrite \({1 \over a} + {1 \over b} \) as \({b+a} \over {ab}\)

Using Vieta's, the sum of the roots (a+b) is \(-b \over a\) and the product (ab) is \(c \over a\)

This means that the sum of the roots is \({4 \over 4} = 1\), and that the product of the roots is \(1 \over 4\)

Substituting this in, we find that \(\large{{1 \over a} + {1 \over b} = {1 \over {1 \over 4}} = \color{brown}\boxed{4}}\)

"The baker in my town has cooked chocolate, nut and butter cookies. 1/5 of the total number of cookies were chocolate, 1/4 of the remaining cookies were nuts and the rest were butter. How many shortbread cookies do I cook for each chocolate chip cookie?"

The question doesn't make sense. What is a shortbread cookie? Is it supposed to be butter?

Pull over terms to the left side:

4x²-4x+1=0

Factor:

(2x-1)²=0

a would be 1/2. b would also be 1/2.

1/a + 1/b would be 4.

2) Because we have a right triangle, we know the circumradius is half of the hypotenuse (longest side length). This means that the circumradius is \(\color{brown}\boxed{6.5}\)

1) Use Heron's formula: \(\sqrt{s(s-25)(s-25)(s-30)}\), where \(s = {{25 + 25 + 30} \over 2}\)

First -- simplify the middle expression:  t - 2 + 7  =  t + 5.

We now have:  t/3 - 5  <  t + 5  <=  - 3t

Divide this into two inequalities:  t/3 - 5  <  t + 5          and          t + 5  <=  -3t

Solve each one independently:    t - 15  <  3t + 15                          4t  <=  -5

                                                        -30  <  2t                                    t  <=  -5/4

                                                         -15  <  t

Combining these answers:       -15  <  t  <=  -5/4

There is a \(6 \over 13\) chance of picking a boy the first time (6 boys over 13 total students)

After that, there is a \(5 \over 12\) chance of picking a boy (5 boys, 1 was chosen, over 12 remaining students)

Thus, the total probability is \({6 \over 13} \times {5 \over 12} = \color{brown}\boxed{5 \over 26}\)

We know that \(\cos = {\text{adjacent} \over \text{hypotenuse}}\)

This means that \(AB = 7\). 

Applying the Pythagorean Theorem, we know that \(AC = \sqrt{51}\).

We also know that \(\tan = {\text{opposite} \over \text{adjacent}}\)

Thus, \(\tan{C} = {7 \over \sqrt{51}} = \color{brown}\boxed{7 \sqrt{51} \over 51}\), just as whyamidoingthis found :)

AB is the adjacent for B, and BC is the hypotenuse. Since AB/BC = AB/10 = 7/10, AB=7.

Using the Pythagorean theorem, you can deduce that AC = \(\sqrt {51}\)

For C, AB is the opposite and AC is the adjacent, so tan C would be \(7\sqrt{51}\over51\).

My trigonometry skills are... lacking, so I would read this over to see if I made any fundamental mistakes.

The probability that the first student chosen is a boy is  1/6.

Now, there are only 5 boys remaining, so the probability that te second student chosen is a boy is  1/5.

Combine these two probabilities by multiplying  (1/6) x (1/5)  =  1/30.

Let \(CE = x\).

By similar triangles, \(BE = {3 \over 8} x\)

Because \(\triangle BEC \) is a right triangle, we can apply the Pythagorean Theorem: \(x^2+{9 \over 64}x^2=9\)

Solving, we find \(x = \sqrt{576 \over 73}\), which can be simplified to \(\color{brown}\boxed{24 \sqrt{73} \over 73}\)