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Vikentia [17]
3 years ago
9

Please answer and if right i will mark brainliest

Mathematics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

Your answer is D.

Step-by-step explanation:

Your answer is D.

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Im not sure how to do this or what function to use SIN, COS, or TAN but if you could walk me through it that would be so so so a
Elanso [62]

Answer: 3.8

Step-by-step explanation:first, you need to figure out what side you’re looking for. Side AC is opposite of the angle you’re given. They also give you length of side AB which is 4. That is the hypotenuse (side directly across from right angle). So, you have the hypotenuse, and need the find the opposite side measure. Which option uses opposite and hypotenuse? That’s Sin. So, you would do: sin70 = ? / 4.

sin (of given angle)= opposite / hypotenuse.

Sin (70 degrees) = ? / 4

Then you solve for “?”

Type sin70 into your calculator, then multiply both sides by 4 to get “?” by itself. You should get 3.758770483, which rounds to 3.8

5 0
2 years ago
Games at a carnival cost $3. prizes awarded to winners cost $145.65. how many games must be played to make $50
stich3 [128]

Answer:

Step-by-step explanation:

3×1405 and $.65

8 0
3 years ago
12x^3-11x^2+9x+18 divided by 4x+3
atroni [7]
12x^3-11x^2+9x+18 divided by 4x+3

put he division into fraction

12x^3-11x^2+9x+18/4 x +3

reduced fraction by 2

12x^3-11x^2+9x+9/2 x +3

calculate sum

12x^3-11x^2+27/2 x +3
that is your answer ^

hope this helps :)
3 0
2 years ago
A man buys 6 ½ of petrol daily. How many litres of petrol does he purchase in six days​
nalin [4]

Answer:

39 liters

Step-by-step explanation:

6.5×6=39

so 39 litres

6 0
2 years ago
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
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