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Asif works for 3 days....   now the job has 7 days remaining for Asif alone

1/7   + 1/x  = 1/5

solve for x = 17.5 days for the slow assistant to finish the remaining job without Asif

Your formulas are not correct Pangolin     

The surface area of a sphere is A pi cm^2 and the volume of the sphere is B pi cm^3 where both A and B are three-digit whole numbers. Find the radius of the sphere. 

I get  r=6  or  r=9

If r= 6  then A=144  and B=288

If r=9  then   A=329     and     B=989

Find how many digits in the base 10 number 9^(8^7) has when written in base nine

there are 5 even numbers, and the answer is 120/55440=60/27720=30/13960=15/6980=1/462.

because base 9 powers have the same number of digits, and 8^7 is 2097152, so there is 2097152 digits.

Symbols and Truth Values Let j represent _July is a warm month._ (True)
Let d represent _I am busy every day._ (False)
Let g represent _I work in my garden._ (True)
Let f represent _I like flowers._ (True)

j = July is a warm month
d = I am NOT busy every day.
g =I work in my garden.
f =I like flowers.

For each compound statement in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the compound statement is true or false.

I am not sure about your notation but this is my assumptions

a. j -> g               July is a warm month so I work in my garden
b. d -> ~g            I am not busy every day so I do NOT work in my garden
c. f -> g                I like flowers so I work in my garden
d. ~g -> ~j            I do NOT work in my garden so I do NOT like flowers.

e. (j _ f) -> d
f. (j _ g) -> f

Maybe 'so' can or should be replaces with some other word, like because, or hence ... 

You will have to tell me what the underscore symbol means.

Thanks Pangolin,

You forgot that the letters and numbers can be reversed so you will need to multiply your answer by 2.

Please proofread your questions and stop wasting everyone's time.

Thanks Ginger,

Unfortunately, I really do not understand the logic behind what you have done.

I roll a die and flip three coins at the same time, and repeat this multiple times. What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

Here is my extremely dubious attempt.

For any individual turn 

   P(HHH)=1/8      P(not HHH)= 7/8

   P(6)=1/6            P(not 6) = 5/6

For any 2 consecutive turns

   P(HHH any number, HHH not 6 (or vise versa)) = 1/8 * 1* 1/8 * 5/6 *2 = 5/192

    P(6  any number of heads, 6 not all heads, (or vise versa)) = 1/6 *1 * 1/6 * 7/8 *2 = 7/144

I am only comparing these two things so the total for comparison is  5/192 + 7/144 = 43/576

Prob of favourable outcomes / prob or relevant outcome = 5/192  divided by  43/576 = 15/43

15/43 is approximately 0.3488

I doubt very much that mine is correct, Ginger seems more confident with her answer.

Solution 1:

Probability of two (2) sixes on a single die occurring before the occurrence of two (2) sets of three heads, in two sequential events.

The first roll: the probability of a six is (1/6). At this point, the outcome of the coins does not matter. Probability: (1/6) * 1 ≈ 0.16667

The second roll: the probability of a six is (1/6). At this point, the outcome of the coins cannot be three heads. Probability: (1/6) * (1-(1/8)) ≈ 0.14583

The two events can occur in any order.  Mean probability (0.16667 + 0.14583)/2   ≈ (0.15625)

----

Probability of two (2) sets of three heads occurring before or in concurrence with two (2) sixes occurring on a single die, in two sequential events.

The first roll: the probability of three heads is (1/8). The outcome of the coins does not matter. Probability: (1/8) * 1 ≈ 0.125

The second roll: the probability of three heads is (1/8). The outcome of the coins does not matter. Probability: (1/8) * 1 ≈ 0.125

The two events are not distinguishable.  Mean probability: (0.125 + 0.125)/2 = 0.125

----------------------------

Solution 2:

Each event can have 48 unique arrangements of die number and ordered (H/T) coin sets. Any of these 48 arrangements can occur twice in the two events.

In the two (2) events, there are 96 total arrangements, and there are 15 arrangements that give success to the die. Probability of success for the die in two events: (15/96) = (0.15625).

In the two (2) events, there are 96 total arrangements, and there are 12 arrangements that give success to the set of coins. Probability of success for coins in two events (12/96) = (0.125)

The question asks:

What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

Rewording this to clarify:

What is the probability that I flip a set of three coins to three heads twice in sequence before I roll a 6 on a die twice in sequence?

In a large sample of twin events 71.875% of the time there is no success for either the die or coins. For 28.125% of the either the coins or the die have a success.   In this space, the coins are successful 44.44% of the time and the die is successful 66.66% of the time. So the probability of set 3 coins has three heads twice in sequence before a die has two sixes in sequence is 44.44%  

Ambiguous wording aside, this is an interesting question.  

GA

--. .-

Suppose g(x)=4x^2+8x+13. Does g have an inverse? If so, find g^{-1}(25). If not, enter "undef".

g(x) only has an inverse if you restrict the domain.

g(x) is a concave up parabola.  At the vertex       \(x= \frac{-b}{2a}=\frac{-8}{8}=-1\)

So if you restrict the domain to x>-1   or   X<-1     then it will have an inverse, otherwise, it does not.

Since no restriction is given, the inverse is undefined.

For the die, think of it as a single six occurring in event one, then a single six occurring in event two.

For the coins, think of it as a set of three heads occurring in event one, then a set of three heads occurring in event two.

The in a row element exists only after two events. 

------

 I am being pedantic with the wording, which is appropriate in mathematics.

Yes, it is. Pedantic wording reduces ambiguity.

My description above provides an interpretation for a solution.

-----

Yes. I’ll post a solution

GA

--. .-

Anyway,  I am being pedantic with the wording, which is appropriate in mathematics.

I can never remember how to do these. Would you like to answer it for us?

Hi Ginger,

^{"Two events are required to meet the criteria.}

^{First event: roll a die and flip three coins at the same time.}

^{Second event: roll a die and flip three coins at the same time.}

^"

^{At this point, it is now possible to have rolled two sixes twice in a row and/or flip three heads twice in a row."}

------------------------

At this point that is not possible.

Two sixes twice in a row = 4 sixes

just as,

3 heads twice in a row = 6 heads.

No problem; I appreciate your answers.

Your answer is more than wrong; it’s atrocious!

GA

--. .-

Hi Melody,

Two events are required to meet the criteria.

First event: roll a die and flip three coins at the same time.

Second event: roll a die and flip three coins at the same time.

At this point, it is now possible to have rolled two sixes twice in a row and/or flip a set of three heads twice on a set of coins two times in a row.

GA

_{Edit: strikeout of redundant word(s) that alters intended meaning of statement.}  

--. .-

A circle has a radius of 25. A circular sector, with an angle of 345.6 degrees at the center, is cut from the circle, and then rolled to form a cone. Find the volume of the cone.

Hello Guest!

\(C_{cone}=2\cdot 25\cdot \pi \cdot \frac{345.6}{360}=\color{blue}150.796\)

\(r_{cone}=\frac{C_{cone}}{2\cdot \pi}=\color{blue}24\)

\(h_{cone}=\sqrt{r_{circle}^2-r_{cone}^2}=\sqrt{25^2-24^2}=\color{blue}7\)

\(V=\frac{1}{3}\cdot \pi \cdot r_{cone}^2\cdot h_{cone}=\frac{1}{3}\cdot \pi \cdot 24^2\cdot 7\)

\(V=4222.3\)

  !

Sorry jugoslav, there was no answer when I started.

I roll a die and flip three coins at the same time, and repeat this multiple times. What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

How can you roll 2 sixes when you only have one die?

\((x-2)^2(x+2)^2\)

The easiest way to do this is to rearrange it as follows.

\((x-2)^2(x+2)^2=(x-2)(x+2)*(x-2)(x+2)\)

that is a good start for you. 

A circle has a radius of 25. A circular sector, with an angle of 345.6 degrees at the center, is cut from the circle, and then rolled to form a cone. Find the volume of the cone.

r = {[(50 * pi / 360) * 345.6] / pi} / 2

r = 24

h = sqrt(25² - 24²) 

h = 7

V = pi*r²*h / 3 = pi*24²*7 / 3 = 4222.3 u³   or   1344pi u³

A rectangular prism 3cm by 5cm by 8cm is painted mauve and is then cut into 1 cm cubes. what is the total unpainted surface area of these cubes ?

Hello Guest!

\(n=\frac{V}{1\cdot cm^3}=\frac{3\cdot 5\cdot 8\cdot cm^3}{1\cdot cm^3}=\color{blue}120\)

\(S_{unpainted}=n\cdot 6\cdot cm^2=120\cdot 6\cdot cm^2=\color{blue}720cm^2\)

\(S_{inside}=S_{unpainted}-S_{outside}=720cm^2-2\cdot (3\cdot 5+3\cdot 8+5\cdot 8)cm^2\)

\(S_{inside}=562cm^2\)

  !

Chance that the dice get all sixes = 6^3, twice in a row = (6^3)^2 = 1/44656

Chance that the coins will get all heads = 2^3, twice in a row = (2^3)^2 = 1/16

So the three coins twice in a row is 2916 times more likely.

Sorry if I got this wrong, my probability isn't very strong.