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Mila [183]
3 years ago
13

Every time kelly eats 3 cookies, Diego can finish 4 cookies. When are they going to eat the same number of cookies?

Mathematics
1 answer:
maw [93]3 years ago
8 0

Answer:

In about 4 hours I think

Step-by-step explanation:

You might be interested in
Based on a kc value of 0.200 and the given data table, what are the equilibrium concentrations of xy, x, and y, respectively?
MariettaO [177]
There is no given data table but based on the question, the reaction is
xy <=> x + y
If we let M as the initial concentration of xy and c as the in the concentration after the dissociation, then we can use the ICE method
    xy   <=>   x    +   y
I    M
C  -c            c         c
-----------------------------
E  M-c          c         c

Solve for c using
Kc = c(c) / (M - c)
And the concentration of the xy, x, and y can then be determined
3 0
3 years ago
Help please!!! photo above
Viktor [21]

Answer:

5<*line under* x >*line under*9

I think

4 0
3 years ago
How many centimeters make up a triangle.
Eva8 [605]

Answer:

120cm

Step-by-step explanation:

A triangle has a perimeter of 120 cm and two sides of 30 cm and 40 cm. Calculates the area. A right triangle has a cathetus of 10 cm and the hypotenuse 26 cm.

3 0
3 years ago
Sam divided a rectangle into 8 congruent
VMariaS [17]

Answer:

40cm²

Step-by-step explanation:

pretty sure bc there are 8 equal 5 cm² pieces meaning you multiply by 8 for the answer. 5•8=40. I think rofl

3 0
3 years ago
A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.
fenix001 [56]

Answer:

1/32

15/32

Step-by-step explanation:

For a fair sided coin,

Probability of heads, P(H) = 1/2

Probability of tails P(T)  = 1/2

For a coin tossed 5 times,

P( All heads)

= P(HHHHH),

= P (H) x P(H) x P(H) x P(H) x P(H)

= (1/2) x (1/2) x (1/2) x (1/2) x (1/2)

= 1/32 (Ans)

For part B, it is easier to just list the possible outcomes for

"at most 2 heads" aka "could be 1 head" or "could be 2 heads"

"One Head" Outcomes:

P(HTTTT), P(THTTT) P(TTHTT), P(TTTHT), P(TTTTH)

"2 Heads" Outcomes:

P(HHTTT), P(HTHTT), P(HTTHT), P(HTTTH), P(THHTT), P(THTHT), P(THTTH), P(TTHHT), P(TTHTH), P(TTTHH)

If we count all the possible outcomes, we get 15 possible outcomes representing "at most 2 heads)

we know that each outcome has a probability of 1/32

hence 15 outcomes for "at most 2 heads" have a probability of

(1/32) x 15  = 15/32

8 0
3 years ago
Read 2 more answers
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