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3 years ago

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sketch a graph of the following - At midnight the water at a particular beach is at high tide. At the same time a gauge at the e

nd of a pier reads 10 feet. Low tide is reached at 6 AM when the gauge reads 4ft.

1 answer:

never [62]3 years ago

5 0

In the attachment is a graph of the cosine function that represents changing of the tide.
y max = 10 ft at midnight
y min = 5 ft at 6 AM

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Help please <br><br>1 thought 3

Anastaziya [24]

A function is a relation that has one output for a given input.

For the first one, there is no one x value with two or more y values so it is a function.

The second example is also a function because a certain value can only have one cube root.

For problem number 3 input "-3" for every instance of x in h(x).

So, h(-3)=2(3^2)-1= 17

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3 years ago

Use long division to convert 36\75 to a decimal

REY [17]

I’ll help....

0.48 is the answer

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3 years ago

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At rehearsal for the school play, jack sings three times as many songs as Jennifer, and then dances to 5 more songs, just to sho

Brilliant_brown [7]

Step-by-step explanation:

3x+5=44

3x=44-5

3x=39

x=39÷3

x=13

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2 years ago

What are the ordered pairs of the

9966 [12]

Start by writing the system down, I will use $y$ to represent $f(x)$

$y=x^2-5x+3\wedge y=-3$

Substitute the fact that $y=-3$ into the first equation to get,

$-3=x^2-5x+3$

Simplify into a quadratic form ( $ax^2+bx+c=0$ ),

$x^2-5x+6=0$

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

$x^2+x(m+n)+mn=0$

which then must factor into

$(x+m)(x+n)=0$

And the solutions will be $m,n$ .

Clearly for small coefficients like ours $5,6$ , this is very easy to figure out. To get 5 and 6 we simply say that $m=3, n=2$ .

This fits the definition as $5=3+2$ and $6=2\cdot3$ .

So as mentioned, solutions will equal to $m=3, n=2$ but these are just x-values in the solution pairs of a form $(x,y)$ .

To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.

So the solution pairs are $(2,0)$ and $(3,0)$ .

Hope this helps :)

5 0

3 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

3 years ago