16 tons 400 pounds divided by 5

Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate

Leona [12618]

Answer: Initially, he had 27 sweets.

Step-by-step explanation: The most logical approach is to work backwards from what remained after the third day to the start of the first day.

On the third day, he consumed one-third of his sweets and was left with 8. If we let the total sweets on day three be denoted as a, then one-third of a equals what he ate and the two-thirds left equals 8, giving us:

8/a = 2/3

By cross-multiplying, we find:

8 x 3 = 2a

Therefore, 24 = 2a

This leads to a = 12.

Let the sweets on day two be represented as b. If he consumed one-third of b and was left with 12, we have the same structure; hence:

12/b = 2/3

Cross-multiplying gives:

12 x 3 = 2b

So, 36 = 2b, leading to b = 18.

Denote the number of sweets on day one as x. If one-third of x was eaten and 18 remained, we can set up the equation:

18/x = 2/3

Again, cross-multiplying results in:

18 x 3 = 2x

Which simplifies to 54 = 2x, yielding x = 27.

Thus, Tim received 27 sweets at the start.

3 0

3 months ago

Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.

AnnZ [12381]

A="b is situated in the center"

B="c lies to the right of b"

C="The letters def occur sequentially in that arrangement"

a) b can occupy 7 positions; however, only one of these is the center. Therefore, P(A)=1/7

b) Let X=i; "b holds the i-th position"

Y=j; "c occupies the j-th position"

$P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}$

P(B)=1/2

c) Let X=i; "d holds the i-th position"

Y=j; "e occupies the j-th position"

Let Z=k; "f is in the i-th position"

$P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}$

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

$P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420$

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

Events A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

Events A and C are not independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

Events B and C are not independent

8 0

2 months ago

Mike is 5 years old and Randy is 8 years old. Select all proportional relationships that have the same ratio to that of Mike and

lawyer [12517]

Response:

Explanatory steps:

I am unsure as well. However, I can state that the first one is

Sarah is 40 years old and her mother is 64 years old.

7 0

2 months ago

What is the solution to the system of equations graphed below? Please Help (:

Zina [12379]

C. (1,5) The solution represents where two lines intersect.

5 0

1 month ago