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Afina-wow [57]
3 years ago
15

When calculating the circumference of a circle, sometimes the radius is given instead of the diameter. What is the relationship

between radius and diameter?
a. radius is twice the diameter
B. diameter is half the radius
C.diameter is twice the radius
D.no relationship, cant find circumstance with radius ​
Mathematics
2 answers:
vaieri [72.5K]3 years ago
4 0

Answer:

Answer: C) diameter is twice the radius

Step-by-step explanation:

edge2021

alex41 [277]3 years ago
3 0
The diameter is double the size of the radius
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Which property is illustrated? 8(a + c) = 8a+ 8c​
alekssr [168]

Answer:

Distributive property

Step-by-step explanation:

Here, we want to evaluate the property illustrated

From what we can see, we have that each of the inner values are multiplied by the outside value

This is simply a display of distributivity

We can conclude that we have a right distributive property in place here

4 0
3 years ago
Find the value of y when x equals 10 5x-y=54
g100num [7]
5x-y=54
x=10
50-y=54
-y=54-50
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7 0
3 years ago
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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
Please help quickly I will give you brainiest But please don't make me report you
svp [43]

Answer:

A.) -4 Yards

B.) 60 Yards

Step-by-step explanation:

Hope this helps you <3

3 0
3 years ago
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Heck did I do wrong!?!?! IXL sucks
zysi [14]

Answer:

∠G, ∠HGI, ∠IGH

Step-by-step explanation:

» <u>Concepts</u>

To name an angle, you MUST include the vertex either in the middle or have it standing by itself. To clarify, you can either call this angle ∠G, ∠HGI, or ∠IGH because they all include the vertex, G, in the middle or by itself. You put ∠HIG, which is incorrect because the vertex <u>is not </u><u>I</u>.

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