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3 years ago

This one please if you can

Mathematics

2 answers:

Deffense [45]3 years ago

7 0

1.Given, x={p,q}.

Now, possible subsets of x are ϕ,{p},{q},{p,q}.

2.if A = fa b, c. d, e and B of a, e, i, o, u. list the elements of B - A and also show it in a venn diagram by

shading

hope it helps

merryyyy Christmassss

happy holidays

have a great day/night

marishachu [46]3 years ago

6 0

You should mark them brainliest

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Anwser the question quickly!

cluponka [151]

B is the correct anwser

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3 years ago

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Let f(x)=(8x^4-3)^3 and g(x)=8x^4-3 given that f(x)=(h g)(x) , find h(x)

Harman [31]

Answer:

$h(x) = x^{3}$

Step-by-step explanation:

$f(x) = (8 \times x^{4}- 3)^{3}$ ------------(1) (given)

$g(x) =(8 \times x^{4}- 3)$ -------------(2) (given)

so,

f(x) = h(g(x)) (given)

= $(g(x))^{3}$ -----(from (1) and (2))

⇒ $h(k) = k^{3}$ -----(taking g(x) = k)

⇒ $h(x) = x^{3}$ is the function.

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3 years ago

ILL GIVE BRAINLIEST! HELP!

KonstantinChe [14]

Answer:

is B))))

Step-by-step explanation:

brainliest pls

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2 years ago

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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) =

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

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3 years ago

4.289 rounded to the nearest tenth

Novay_Z [31]

Answer:

4.3

Step-by-step explanation:

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2 years ago

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This one please if you can​

This one please if you can