All categories

3 years ago

The product of 9 and the quantity 5 more than a number t is less than 5

1 answer:

3 0

$9(t+5) \ \textless \ 5\ \ \ |use\ distributive\ property:a(b+c)=ab+ac\\\\(9)(t)+(9)(5) \ \textless \ 5\\\\9t+45 \ \textless \ 5\ \ \ |subtract\ 45\ from\ bot\ sides\\\\9t \ \textless \ -40\ \ \ \ |divide\ both\ sides\ by\ 9\\\\\huge\boxed{t \ \textless \ -\dfrac{40}{9}\to t \ \textless \ -4\dfrac{4}{9}}$

You might be interested in

Leo made a container to store his camping gear. The container is in the shape of a rectangular prism. The container has material

Contact [7]

(a) What is the capacity of the container?
Volume = length * width * height
V = 4ft * 2ft * 3ft
V = 24 ft³

(b) How much material was used to make the container?
Surface area = sum of 6 faces of the prism
4 ft x 3 ft = 12 ft² → 12 ft² x 2 = 24 ft²
4 ft x 2 ft = 8 ft² → 8 ft² x 2 = 16 ft²
2 ft x 3 ft = 6 ft² → 6 ft² x 2 = 12 ft²

Surface area = 24 ft² + 16 ft² + 12 ft² = 52 ft²

4 0

3 years ago

Read 2 more answers

Please help me someone ASAP

Artyom0805 [142]

The Answer is b. Hope that helps

5 0

3 years ago

In a jar of coins, the ratio of nickels to dimes is 2:7. If the jar contains 350 nickels, how many dimes are there? A) 175 B) 35

Dvinal [7]

No nickles = 2x and that of dimes = 7x so 2x = 350 get x and the no of dimes
so like 2x = 350 divide by 2 on both sides so x = 175 ... and no of dimes = 7*x = 7*175 = 1225
Hope it helps :)

3 0

3 years ago

Read 2 more answers

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$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$