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3 years ago

Can someone help me i really needed

Mathematics

2 answers:

romanna [79]3 years ago

8 0

Answer:

C: √74

Step-by-step explanation:

We have to use the Pythagorean theorem to answer this (a² + b² + c²). We have a and b so all we have to do is square 5 and 7 (a and b) so we get a² b², then we add this to get c² as stated by the Pythagorean theorem. So, 7² (7 x 7) is 49, and 5² (5 x 5) is 25. 25+49 (a² + b²) is 74 (c²). So the square root of this would be the absolute length of the hypotenuse because this is the hypotenuse squared. Meaning that the answer would be √74 (square root of 74)

AlladinOne [14]3 years ago

5 0

The answer would be the third option or C

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URGENT GEOMETRY!!

Darina [25.2K]

Answer: 140 degrees

A straight line is 180 degrees. Since we know one of the angles in the line is 40, we just subtract 40 from 180. So, the measure of angle 1 plus the measure of angle 2 is 140 degrees.

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2 years ago

Estimate the perimeter of the figure to the nearest whole number. PLZ SOLVE PLZ

soldier1979 [14.2K]

Answer:

The perimeter is approximately 19.

Step-by-step explanation:

Three of the sides are roughly four segments long and two are around three and a half.

4 + 4 + 4 + 3.5 + 3.5 = 19

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3 years ago

Read 2 more answers

Write an exponential function in the form y = ab that goes through points (0,19)<br>and (5,4617).

gtnhenbr [62]

Answer:

y = 1/2 (2)x

Step-by-step explanation:

Using the equation y = abx , substitute both of your given points into that equation.

2 = ab2 and 4 = ab3 Solve each equation for a.

2⁄b2 and 4⁄b3 = a Therefore, 2⁄b2 = 4⁄b3

Cross multiply: 2b3 = 4b2 Divide both sides by b2

2b = 4 a = 2/4 = 1/2

b = 2

3 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

A fair four-sided spinner has the numbers 1 to 4 marked on its sides. It is spun twice and the two scores are added together.

scZoUnD [109]

Answer:

Probability of getting doubles = 1/4

Step-by-step explanation:

Given - A fair four-sided spinner has the numbers 1 to 4 marked on its sides. It is spun twice and the two scores are added together.

To find - Using a sample space diagram, find the probability of getting a double.

Proof -

Given that,

A four-sided spinner is spin twice.

So,

The Sample Space becomes

S = { (1, 1), (1, 2), (1, 3), (1, 4)

(2, 1), (2, 2), (2, 3), (2, 4)

(3, 1), (3, 2), (3, 3), (3, 4)

(4, 1), (4, 2), (4, 3), (4, 4) }

So,

n(S) = 16

Now,

Let A is the event Getting a double, So,

A = { (1, 1), (2, 2), (3, 3), (4, 4) }

and

n(A) = 4

So,

The Probability of getting doubles = n(A) ÷ n(S)

= 4 ÷ 16

= 1/4

∴ we get

Probability of getting doubles = 1/4

6 0

3 years ago