A single die is rolled twice. Find the probability of rolling an Odd number the first time and a number greater than 4 the second time.

Answers

Answer 1

Answer:

The probability of rolling an odd number the first time and a number greater than 4 the second time is 1/6

The sample space of a single die is

$\mathbf{S = \{1,2,3,4,5,6\}}$

So, the total sample is 6

The odd numbers are

$\mathbf{Odd = \{1,3,5\}}$ --- 3 odd numbers.

So, the probability of selecting an odd number is:

$\mathbf{P(Odd) = \frac 36}$

Simplify

$\mathbf{P(Odd) = \frac 12}$

The numbers greater than 4 are

$\mathbf{Greater = \{5,6\}}$ --- 2 numbers greater than 4.

So, the probability of selecting a number greater than 4 is:

$\mathbf{P(Greater) = \frac 26}$

Simplify

$\mathbf{P(Greater) = \frac 13}$

The probability of rolling an odd number the first time and a number greater than 4 the second time is calculated as follows:

$\mathbf{P = P(Odd) * P(Greater)}$

So, we have:

$\mathbf{P = \frac 12 * \frac 13}$

$\mathbf{P = \frac 16}$

Hence, the probability is 1/6

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Find the coordinates of C' after a reflection across the liney = 1 and then x = 2. Write in the form (a,b)Part 5a

A triangle has squares on its three sides as shown below what is the value of x

Answers

Answer:

5 cm

Step-by-step explanation:

Suppose that the population of deer in a state is 1,500 and is growing 2% each year. Predict the population after 4 years.

Answers

Required number
= 1500*(1+2%)^4
=1624(cor.to.the nearest integer)

Evaluate $\rm (3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+\ldots+(101^2+1)/(101^2-1) =$
With step by step explanation !

Answers

It's easier to deal with the symbolic sum (in sigma notation),

$\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1)$

Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives

$((2k+1)^2+1)/((2k+1)^2-1) = (4k^2 + 4k + 2)/(4k^2 + 4k)$

$=\frac12*(2k^2 + 2k + 1)/(k^2 + k)$

$=\frac12\left(2+\frac1{k(k+1)}\right)$

$=\frac12\left(2 + \frac1k - \frac1{k+1}\right)$

and it's the latter two terms that reveal a telescoping pattern.

In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that

$\frac1{k(k+1)}=\frac ak+\frac b{k+1}$

$1 = a(k+1) +bk =(a+b)k+a$

which gives a = 1, and a + b = 0 so that b = -1.

Our sum has been rearranged as

$\displaystyle\frac12\sum_(k=1)^(50)\left(2+\frac1k-\frac1{k+1}\right)=\sum_(k=1)^(50)1+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)=50+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)$

The remaining telescoping sum is

1/2 [(1/1 - 1/2) + (1/2- 1/3) + (1/3- 1/4) + … + (1/48- 1/49) + (1/49- 1/50) + (1/50 - 1/51)]

and you can see how there are pairs of numbers that cancel, so that the sum reduces to

1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51

So, our original sum ends up being

$\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1) = 50 + (25)/(51) = \boxed{(2575)/(51)}$

The product of (a + b)(a − b) is a perfect square trinomial.
A. Sometimes b. Always c. Never

Answers

never
perfect squre trinomial is in form
(a+b)(a+b) or (a-b)(a-b)
that is not equal to (a+b)(a-b)

never

Answer:

C. Never

Step-by-step explanation:

If f(x) = 3x + 2 and g(x) = x2 + 1, which expression is equivalent to (fxg) (x)?

Answers

Answer:

$(f* g) (x)=3x^3+2x^2+3x+2$

Step-by-step explanation:

Given : If $f(x) = 3x + 2$ and $g(x)=x^2+1$

To find : Which expression is equivalent to $(f* g) (x)$ ?

Solution :

We can write,

$(f* g) (x)=f(x)* g(x)$ ....(1)

We know, $f(x) = 3x + 2$ and $g(x)=x^2+1$

Substituting the values in (1),

$(f* g) (x)=(3x+2)* (x^2+1)$

Multiply term by term,

$(f* g) (x)=3x^3+3x+2x^2+2$

$(f* g) (x)=3x^3+2x^2+3x+2$

Therefore, The expression is equivalent to $(f* g) (x)=3x^3+2x^2+3x+2$

$f(x)=3x+2\nThe\ domain\ D_f=\mathbb{R}\n\ng(x)=x^2+1\nThe\ domain\ D_G=\mathbb{R}\n\nD_f=D_g\ therefore\ (f* g)(x)=f(x)* g(x)\n\n(f* g)(x)=(3x+2)(x^2+1)=3x^3+3x+2x^2+2=\boxed{3x^3+2x^2+3x+2}$

Brandon is 6 times as old as Cora. In 4 years, Brandon will be only twice as old asCora will be then. Find Brandon’s age now.

Answers

$C-Cora's\ age\ now\nB-Brandon's\ age\ now\n\nB=6C\ \ \ and\ \ \ B+4=2(C+4)\n\n6C+4=2C+8\n6C-2C=8-4\n4C=4\ \ \ \Rightarrow\ \ \ C=1\ \ \ \Rightarrow\ \ \ \ B=6\cdot1=6\n\nAns.\ Brandon\ is\ now\ 6\ years.$

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