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Bad White [126]
3 years ago
8

Can someone actually help me with this? I got so many fake answers :( Math Pls :') <3

Mathematics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

I saw this question wasn't answered by anyone yet so I can help you with this one!

A is 8

B is 4

C is 3

D is 5

Area: 100

Now to find area with these its the same except for the triangles here in this photo I showed the area for each figure BY ITSELF so the thing with the triangles the formula is to do base × height thennn ÷ by 2 So i did 4 × 3 = 12 then divide by 2 and got 6 for both of those triangles as you see in the photo! Then I just did base × height for the rectangles! I did 8 × 4 = 32 and since it had a matching rectangle there was another 32 as displayed in the photo then the middle triangle I simply just did 8 × 3 cause thats the base and height and I got 24 now we just have to add these up! 6 + 6 + 32 + 32 + 24 = 100 so 100 is our area! Hope that makes sense and an easy trick for those triangles to remember to divide by 2 is a triangle is half of a rectangle/square like I displayed in another photo Hope that helps!

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The answer is B since the equation consists of multiplication. B has Michael buying 7 pass for 5.50 each. A has Marcos buying 7 candies for a total of 5.50 meaning that they used addition. B used multiplication for the 5.50 (tickets each) x 7 (ticket number).

So the answer is B :D

Thank you :DDDD

4 0
3 years ago
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Carlota bought 5 slices of pizza and h hamburgers. The cost of a slice of pizza is $p and a hamburger is $3 more than a slice of
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During which time interval is the squirrel traveling at the fastest speed?
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3. For which of the following samples would it be appropriate to use t-procedures for inference for the population mean?
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4 0
3 years ago
Exercise 5.2. Suppose that X has moment generating function
soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Step-by-step explanation:

Given:

moment generating function  of X as:

MX(t) = \frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}

a)  Now

Mean, E(X) = M_{X}'(t=0)

Thus,

M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}

or

M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}

also,

E(X^{2})=M_{X}''(t=0)

Thus,

M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}

or

M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}

Therefore,

Mean, E(X) = M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}

or

Mean, E(X) = - 0.5

and

E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}

or

E(X^{2}) = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Therefore,

at x = 0, P(x) = \frac{1}{2}

at x= - 4 ,P(x) = \frac{1}{3}

at x = 5, P(x) = \frac{1}{6}

Thus,

E(X) =\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})

or

E(X) = - 0.5

also,E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})

E(X^{2})  = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0
4 years ago
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