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2 years ago

In a certain isosceles triangle, the base is 1 1/2 times as long as each leg and the perimeter is 63. How long is the base?

Mathematics

1 answer:

son4ous [18]2 years ago

6 0

Answer:

The base is 27 units long.

Step-by-step explanation:

This is a solve the system question. Call the length of the base B and the length of each leg L. We are told:

$B=1.5L$

and

$B+2L=63$

Sub the first equation into the second:

$1.5L+2L=63\\3.5L=63\\L=18$

Now that we have the leg length, sub that into either of the original equations to find the base. I'll use the first equation:

$B=1.5(18)\\B=27$

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. Laura and Alicia both exercise 5 days a week. Laura exercises for 30 minutes each day. Alicia exercises for 45 minutes each da

statuscvo [17]

Answer: 1 hour 15 minutes

Step-by-step explanation:

From the question, we are informed that Laura and Alicia both exercise 5 days a week and that Laura exercises for 30 minutes each day. For the 5 days, she'll exercise for:

= 5 × 30 minutes

= 150 minutes

= 2 hours 30 minutes

Alicia exercises for 45 minutes each day. Fir the 5 days, she'll exercise for:

= 5 × 45 minutes

= 225 minutes

= 3 hours 45 minutes

We then calculate the difference in their exercise per week which will be:

= 3 hours 45 minutes - 2 hours 30 minutes

= 1 hour 15 minutes

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2 years ago

This is to help my wonderful mother. What is 1/3rd of 4 1/2s and 8 1/4s?

crimeas [40]

4 1/2 = 9/2
8 1/4 = 33/4

1/3 x 9/2 = 3/2
1/3 x 33/4 = 33/12

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3 years ago

Which equation represents the circle shown in the graph?

marta [7]

Answer:

i think 3

Step-by-step explanation:

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3 years ago

A mousetrap car is filmed as it runs the racecourse. The film includes 1,200 photo frames taken in 8 seconds. How many frames we

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Answer: 150 frames were taken.

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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

2 years ago