# Posts by Pendergast

Total # Posts: 33

**math 3**

4/10 chance of rolling 10 3/10 chance of rolling 5 2/10 chance of rolling 25 1/10 chance of rolling 27 Expected value is the sum of the outcomes multiplied by probabilities = (4/10)(10) + (3/10)(5) + (2/10)(25) + (1/10)(27)

**math 3**

duplicate question

**math**

sum of first 5 natural numbers = (5)(6)/2 sum of first 21 natural numbers = (21)(22)/2 sum of first 30 natural numbers = 465 = (30)(31)/2 sum of first n natural numbers = (n)(n+1)/2 -- provable by induction

**math**

C(7,2)*C(8,3) ways of choosing 2 out of 7 men and 3 out of 8 women. C(n,k) = n!/[(k!)(n-k)!] C(15,5) ways of picking 5 committee members out of 15 probability of selecting 2 men and 3 women is good over total -- C(7,2)*C(8,3)/C(15,5)

**math**

Probability of getting 0 heads is 1/2^6 Probability of getting at least 1 head is the complement -- 1 - 1/2^6

**math**

Let b be the number of boys, g be the number of girls. "54 pupils in class" b + g = 54 "3 more boys than girls" b = g+3

**algebra**

87/400

**Math Algebra**

There are 3 terms, so it is a polynomial. The only variable is x. The highest power of x is x^2 - the degree is 2.

**math**

The candle has burned up (8hrs)(0.4cm/hr).

**precalculus/trigonometry**

Intersections with y=x^2 at x=-2 and x=4 describes two points: (-2, y=(-2)^2) and (4, y=4^2) Find the line between those two points.

**Calc**

Break it into two integrals, integral (sec^2(x)tan(x))dx and integral(-tan(x))dx. Both can be solved using u substitution.

**pre algebra homework check**

Correct

**pre algebra check**

Correct

**pre algebra homework check**

Correct

**pre algebra**

Does the second roll depend on the outcome of the first roll? (If so, we call the events "dependent". If not, we call them "independent")

**pre algebra answer check**

Correct

**pre algebra check**

Correct

**pre algebra**

The area is A = pi*r^2 We know A = 3.7994ft^2, we know the value for pi. 3.7994ft^2 = pi*r^2 Solve for r.

**Math**

The cliff makes a right angle with the ocean. The two legs are 50m (the cliff) and 950m (distance on the ocean from the cliff to the ship). Draw the triangle for that, the rest follows.

**algebra**

Use long division.

**Algebra 2**

The probability of drawing an even chip is 2/5, and the probability of drawing an odd chip (which does not change due to replacement) is 3/5. (2/5)*(3/5)

**math**

Assuming fair coins, the probability of obtaining heads on one coin is 0.5. Obtaining two heads has probability 0.5*0.5.

**math**

There are 7 bronze trophy winners and 7+14+10=31 classmates in total. 7/31

**Math**

f: T->S is a surjection, so a function g: S->T exists. g takes elements of S to t_i, i in N injectively since f exists. Intuitively g can be rewritten to take elements of S to N. Formally, let h:T->N, with h(t_i) = i. Then h(g): S->N.

**Precalculus check answers help!**

1), 3) check your algebra 4) I'm not sure where you went wrong on this one. However, sqrt(6)-sqrt(2)/4 is greater than 1 (2.0959 in fact) 2), 5) are correct

**geometry**

I assume the the only forces on the rocket after launch are due to gravity. Gravity exerts an acceleration of -32ft/s^2. The height function h(t), derived from acceleration a(t) = -32ft/s^2, is h(t) = (1/2)(-32ft/s^2)*t^2 + v_0*t + h_0. v_0 is the initial velocity, and h_0 is ...

**geometry**

I don't understand the question. You want to find a and b, yet you state their values.

**3rd grade math**

Calm down, it's not that difficult. Your child is making valuable use of his last week in school. Think back to your mathematics background - I'm sure you understand this sufficiently to explain it. 86/9 -- 9*9 = 81, so 86/9 = 9 remainder 5. Find the largest multiple ...

**pre algebra**

There are 4 friends - 4 choices for the first friend to be picked up. For every choice of the first picked up, there are 3 friends left - 3 choices for the second friend to be picked up. Following the pattern, 2 choices for the third friend, 1 choice for the fourth. 4*3*2*1 = 24

**3rd grade math**

Sam has 4 chores to complete - he can choose the order. He has 4 choices for the first chore. For each choice of the first, he can choose 1 of the 3 remaining for the second chore. etc. 4*3*2*1 = 24 This is more generally known as the ordering problem -- there are n! ways to ...

**pre algebra**

The volume of a cone is given by V = (1/3)*pi*r^2*h. We're given V and h, so solve for r. Remember, the diameter is two times the radius.

**pre algebra check**

Correct.

**Geometry**

The 3-dimensional box cannot be described by 2 dimensions.