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3 years ago

A bag contains 90 marbles. some green and some transparent. The ratio of green

Mathematics

1 answer:

Archy [21]3 years ago

8 0

Answer:

54 green marbles

Step-by-step explanation:

So green : transparent = 3:2

so 3+2=5 and 3*90/5 = 54 green marbles

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How do you solve this ?

ss7ja [257]

Answer:

q=7.7

Step-by-step explanation: This is trigonometry and you would use cosine since you have H (29) as your known and A (q) as your unknown. You would do cos(39)29=7.7

4 0

4 years ago

Let Y1, Y2, . . . , Yn be independent, uniformly distributed random variables over the interval [0, θ]. Let Y(n) = max{Y1, Y2, .

Anettt [7]

Answer:

a) $F(y) = 0, y$

$F(y) = \frac{y}{\theta} , 0 \leq y \leq \theta$

$F(y)= 1, y>1$

b) $f_{Y_{(n)}} = \frac{d}{dy} (\frac{y}{\theta})^n = n \frac{y^{n-1}}{\theta^n}, 0 \leq y \leq \theta$

$f_{Y_{(n)}} =0$ for other case

c) $E(Y_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1}= \theta [\frac{n}{n+1}]$

$Var(Y_{(n)}) =\theta^2 [\frac{n}{(n+1)(n+2)}]$

Step-by-step explanation:

We have a sample of $Y_1, Y_2,...,Y_n$ iid uniform on the interval $[0,\theta]$ and we want to find the cumulative distribution function.

Part a

For this case we can define the CDF for $Y_i$ , $i =1,2.,,,n$ like this:

$F(y) = 0, y$

$F(y) = \frac{y}{\theta} , 0 \leq y \leq \theta$

$F(y)= 1, y>1$

Part b

For this case we know that:

$F_{Y_{(n)}} (y) = P(Y_{(n)} \leq y) = P(Y_1 \leq y,....,Y_n \leq y)$

And since are independent we have:

$F_{Y_{(n)}} (y) = P(Y_1 \leq y) * ....P(Y_n \leq y) = (\frac{y}{\theta})^n$

And then we can find the density function calculating the derivate from the last expression and we got:

$f_{Y_{(n)}} = \frac{d}{dy} (\frac{y}{\theta})^n = n \frac{y^{n-1}}{\theta^n}, 0 \leq y \leq \theta$

$f_{Y_{(n)}} =0$ for other case

Part c

For this case we can find the mean with the following integral:

$E(Y_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y y^{n-1} dy$

$E(Y_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^n dy$

$E(Y_{(n)}) = \frac{n}{\theta^n} \frac{y^{n+1}}{n+1} \Big|_0^{\theta}$

And after evaluate we got:

$E(Y_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1}= \theta [\frac{n}{n+1}]$

For the variance first we need to find the second moment like this:

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^2 y^{n-1} dy$

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^{n+1} dy$

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \frac{y^{n+2}}{n+2} \Big|_0^{\theta}$

And after evaluate we got:

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+2}}{n+2}= \theta^2 [\frac{n}{n+2}]$

And the variance is given by:

$Var(Y_{(n)}) = E(Y^2_{(n)}) - [E(Y_{(n)})]^2$

And if we replace we got:

$Var(Y_{(n)}) =\theta^2 [\frac{n}{n+2}] -\theta^2 [\frac{n}{n+1}]^2$

$Var(Y_{(n)}) =\theta^2 [\frac{n}{n+2} -(\frac{n}{n+1})^2]$

And after do some algebra we got:

$Var(Y_{(n)}) =\theta^2 [\frac{n}{(n+1)(n+2)}]$

3 0

4 years ago

Calculate the measure of ∠ PMU and ∠ UPM. m∠ PMU = m∠ UPM =

Harrizon [31]

Answer:

you want to find angle PMB to find PMU

180-21-35= 124

so now, you subtract 180-124 to find PMU

180-124= 56

Now, subtract 56 and 62 from 180 to find angle UPM

180-56-62= 62

m< PMU = 56

m<UPM = 62

4 0

3 years ago

Read 2 more answers

Consider the following sequence and type the next number in the sequence. 55, 48, 41,34

astraxan [27]

Answer:

27 20 13

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

X+2y=10. x+y=6 SOLVE for the value of x and y.

Ksivusya [100]

X+2y=10 ----Eq.1
x+y=6---------Eq.2
from Eq2.
x=6-y
Subtituting the value of X in the first equation
6-y+2y=10
6+y=10
y=4
Plug y=4 in equation 2.
x+4=6
X= 2

7 0

3 years ago