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Basile [38]
4 years ago
12

Write down in terms of n, an expression for the nth term

Mathematics
1 answer:
denpristay [2]4 years ago
3 0

Answer:

a: -3n+1

b:-5n-2

Step-by-step explanation:

a: it is -3+1 because you can see that the sequence is going down by 3, so it is the 3 times table, but because the number is 4, and 4 is 1 more than 3, you put plus one at the end. :)

b: similar to the last, it's going down by five, but it is two less this time, so you put minus two instead.

Hope this helps :),

Lily

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What is true about the dilation?
Lana71 [14]

Answer:

- It is a reduction with a scale factor between 0 and 1.

- It is an enlargement with a scale factor greater than 1.

Step-by-step explanation:

1. Dilation is defined as a transformation of an image in which the size on this image changes but its shape does not change.

2. In dilation, to obtain the new dimensions of the image you must multiply the dimensions of the original image by a number called "Scale factor".

3. When the scale factor is greater than 1 it is an enlargement and when this is between 0 and 1, it is a reduction.  

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7 0
4 years ago
A bag contains 5 blue marbles 3 yellow marbles and 2 red marbles whats the probability of drawing blue marble?
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a

Step-by-step explanation:

a

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3 years ago
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Express the quantity of the Cayennepepper im lemon juice in form one​
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HUH

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3 years ago
Use the function rule. Find y for x = 1, 2, 3, and 4. y = x – 5
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Answer:

is the fourth one

Step-by-step explanation:

just replace x with the numbers given

and then work it out.

y=x-5

y=1-5=-4

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Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j
Elina [12.6K]

Answer:

f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x

E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \

Step-by-step explanation:

We have two random variables X and Y. X \sim Unif(0,2) and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

f_X (x) =\frac{1}{2} , 0\leq x\leq 2

f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x

And on this case we can find the joint density with the following formula:

f(x,y) = f_{Y|X}(y|x) f_X (x)

And multiplying the densities we got this:

f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x

Now with the joint density we can find the expected value E(Y|x) with the following formula:

E(Y|x) = \int y f_{Y|X}(y|x)dx

And replacing we got:

E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \

5 0
4 years ago
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