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tankabanditka [31]
3 years ago
14

Sub to bionic pleasssssse

Mathematics
2 answers:
notka56 [123]3 years ago
8 0

Answer:

BRO HE IS THE BEST SO I SUBBED

BlackZzzverrR [31]3 years ago
4 0
Alright I got you I’ll sub
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What is the simplified form of 15 x to the eighth power over 24 y to the fifth power divided by 4 x to the fourth power over 8 s
IRISSAK [1]
The exponent symbol when typing is ^. so I so rewrite this for you.
(15x^8 / 24y^5) / (4x^4 / 8y^2)

I assumed you meant 8y squared because you said 8 squared.

now when we divide by a fraction, we can actually multiply by it's reciprocal and get the same thing. The word reciprocal really just means flip it over. so:

(15x^8 / 24y^5) * (8y^2 / 4x^4)

let's reduce the coefficients ( the numbers in front of the x and y) to make it easier.

We have 15/24 which can be reduced to 5/8
and 8/4 which is 2/1

so:
(5x^8 / 8y^5) * (2y^2 / x^4)

multiply numerator and denominator
(10x^8y^2) / (8x^4y^5)

now reduce coeffient. 10/8 is 5/4
reduce x: x^8/x^4 is x^4
reduce y: y^2/y^5 is 1/y^3

now put together
5x^4 / 4y^3

so C
3 0
3 years ago
Please help me with geometry :(
eimsori [14]

Answer:

I think it's C is true but make sure

3 0
2 years ago
Read 2 more answers
Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g
Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

    $E[g(x)] = \sum g(x)p(x)$

So,  $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

       $g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also,  $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

   $g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0
2 years ago
Find x and y intercepts for 3x=y+1
Verizon [17]
3x=2y that is the answer
4 0
3 years ago
Read 2 more answers
Please help please please
Elena L [17]

Answer:

x= 11.04

................

6 0
3 years ago
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