The graph is A. Neither even nor odd B.odd C.even D. Symmetrical about the y axis

Mathematics

1 answer:

dybincka [34]3 years ago

3 0

Answer:

Odd

Step-by-step explanation:

The function is odd because cubic functions are always odd.

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A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang

laiz [17]

Answer-

a. Probability that three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

Solution-

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

$\binom{19}{3} = \frac{19!}{3!16!} = 969$

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

$\binom{33}{6} = \frac{33!}{6!27!} =1107568$

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

$\binom{52}{9} = \frac{52!}{9!43!} = 3679075400$

∴P(3 white candies) = $\frac{1073233392}{3679075400} =0.29$

Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

$\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}$

$=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})$

$=(969)(45)(7)(5)(15)=22892625$

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

$\binom{52}{9}=\frac{52!}{9!43!} =3679075400$

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

$\frac{22892625}{3679075400} = 0.006$

6 0

4 years ago

What is bigger 1/3 or 3/5

raketka [301]

$Method\ 1:$

$\dfrac{1}{3}=\dfrac{1\cdot5}{3\cdot5}=\dfrac{5}{15}\\\\\dfrac{3}{5}=\dfrac{3\cdot3}{5\cdot3}=\dfrac{9}{15}\\\\\dfrac{3}{15} < \dfrac{9}{15}\ therefore\ \dfrac{3}{5}\ is\ bigger$