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2 years ago

PLEASE I NEED HELP I GIVE BRAINLIEST

Mathematics

2 answers:

Sonja [21]2 years ago

6 0

Answer:

this has to be a meme

Step-by-step explanation:

are you trolling us

inn [45]2 years ago

6 0

Answer:

A.) $|r| = 2.665$ km

B.) $14$ ° north of east.

Step-by-step explanation:

From these clues, you know that you are taken along the following course: 50 km/h for 2.0 min, turn 90° to the right, 20 km/h for 4.0 min, turn 90° to the right, 20 km/h for 60 s, turn 90° to the left, 50 km/h for 60 s, turn 90° to the right, 20 km/h for 2.0 min, turn 90° to the left, 50 km/h for 30 s.

The car is in the J-direction, the Brainly symbols won't let me insert me math, so yeah.

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MY PFPPPPPPPPPPPPPPPPPPPPPP WHAT HAPPENED!???

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A culture of bacteria doubles every hour. If there are 500 bacteria at the beginning, how many bacteria will there be after 24 h

Mama L [17]

<span>8388608000. you will have to keep multiplying 2 24 times.</span>

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3 years ago

Suppose sin(A) = 1/4. use the trig identity sin^2(A)+cos^2(A)=1 to find cos(A) in quadrant II. around to ten-thousandth.

larisa86 [58]

In quadrant II, $\cos(A)$ will be negative. So

$\sin^2(A) + \cos^2(A) = 1 \implies \cos(A) = -\sqrt{1 - \sin^2(A)} = -\dfrac{\sqrt{15}}4 \approx \boxed{-0.9682}$

8 0

2 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

Need help with a math problem

Greeley [361]

The answer would be $159.85

6 0

3 years ago