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4 years ago

The first 5 multiples of 18 after 0 are?

Mathematics

2 answers:

barxatty [35]4 years ago

8 0

Answer:

18,36,54,72,90

Step-by-step explanation:

ololo11 [35]4 years ago

4 0

Answer:

18,36,54,72,90 are first five multiples of 18

Step-by-step explanation:

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I recipe requires 1/2 cup of flour to make 9 cupcakes using this recipe how much flour is needed to make 12 cupcakes

konstantin123 [22]

Answer:

12/18 cups or 2/3 cups

Step-by-step explanation:

1/2 * 9 = 9/18

1/18 cup to make one cupcake

4 0

3 years ago

Solve the following using one variable in your equation: Water is being drained out of a tank through 2 pipes at the rate of 330

Vsevolod [243]

Answer:

One pipe drains 140 L/min and the other pipe drains 190 L/min

Step-by-step explanation:

Assume that the pipe which releases less is x L/min

∵ One pipe releases x L/min

∵ Other pipe releases 50 L/min more than it

∴ The other pipe releases x + 50 L/min

∵ Water is being drained out of a tank through these 2 pipes

∵ Water is being drained at rate 330 L/min through them

- Add their rates and equate the sum by 330

∴ x + x + 50 = 330

- Add the like terms in the left hand side

∴ 2x + 50 = 330

- Subtract 50 from both sides

∴ 2x = 280

- Divide both sides by 2

∴ x = 140

∵ x represents the rate of one pipe and x + 50 represents the

rate of the other pipe

∴ One pipe drains 140 L/m

∴ Other pipe drains 140 + 50 = 190 L/min

5 0

3 years ago

I don’t really know what to do.

iragen [17]

If you mean stretch, 1 is 41/4ft and 2 should be 47.63m

6 0

3 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

3 years ago

Evaluate the expression |2x – 5| for x = –3 and for x = 3.

stepladder [879]

Answer:

C. 11, 1

Step-by-step explanation:

I could have sworn I already answered this. Oh well. I guess I can answer it again. Simply plug in -3 and 3 for x, then find the absolute value [ALWAYS POSITIVE].

3 0

3 years ago

The first 5 multiples of 18 after 0 are?​

The first 5 multiples of 18 after 0 are?