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3 years ago

What is the 6th term of the geometric sequence shown?

Mathematics

1 answer:

solniwko [45]3 years ago

8 0

Answer:

Step-by-step explanation:

subtract 20 from 40 it's Zero

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A company uses paper cups shaped like cones for its water cooler. Each cup has a height of 12 cm, and the base has a radius of 5

Alex

Answer:

16.65 cups

Step-by-step explanation:

Step one:

height =12 cm, and

the base has a radius =5.5 cm

Volume of the cup = πr^2h

substitute we have

volume= 3.142*5.5^2*12

volume= 3.142*30.25*12

volume=1140.546cm^3

Step two:

volume of cooler= 18,997 cm^3

hence the number of cups to fill it is

=18,997/1140.546

=16.65 cups

4 0

3 years ago

Why road man holes build in circul type

garik1379 [7]

Answer: where is the question

Step-by-step explanation:

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

⇒ $=\sum e^{tx}p(x)$

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

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,

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

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

also,

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

Thus,

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

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

Therefore,

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

Mean, E(X) = - 0.5

and

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

$E(X^{2})$ = 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,

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

⇒ $=\sum e^{tx}p(x)$

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

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

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

Darius and Anita took the same test. Darius answered 5/6 of the questions correctly, and Anita answered 6/7 correctly. Who got t

RideAnS [48]

Anita because you have to remember fractions are just division problems, 5÷6=.83 repeating (3 is repeating) and 6÷7= about .86

3 0

3 years ago

Read 2 more answers

A plane left 30 minutes late than its schedule time and in order to reach the destination 1500 km away in time it had to increas

schepotkina [342]

Let x= the usual speed
1500/x would equal how long it usually took the plane to read the destination
1500/(x+100) would equal how long it would take the plane now
So $\frac{1500}{x} - \frac{1500}{x+100} = \frac{1}{2}$ because it would lose 30 minutes already
Simplify:
$\frac{1500(x+100)-1500x}{x(x+100)} = \frac{1}{2}$
$\frac{1500x+150000-1500x}{x^2+100x} = \frac{1}{2}$
$\frac{150000}{x^2+100x} = \frac{1}{2}$
$\frac{300000}{x^2+100x} =1$
$300000= x^{2} +100x

x^2+100x-300000=0

x^2+600x-500x-300000=0

x(x+600)-500(x+600)

(x-500)(x+600)

x=500, x \neq -600
$

The usual speed was 500 hm/hr

7 0

3 years ago