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3 years ago

You are dealt one card from a standard 52-card deck. Find the probability of being dealt a six.

Mathematics

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

1/13 ≈ 7.69%

Step-by-step explanation:

There are 4 sixes in a standard deck. So the probability is 4/52 = 1/13, or about 7.69%.

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Find the minimum of the data set.

Natalija [7]

Answer:

23,600.

Step-by-step explanation:

23|6

7 0

3 years ago

A car depreciated $1000 the first year it was owned and driven. In years 2 and

Sidana [21]

AAnswer:

Step-by-step explanation:

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3 years ago

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Identify the sequence graphed below and the average rate of change from n = 1 to n = 3.

dsp73

Answer:

The required sequence is $a_n=20(\frac{1}{2})^{n-1}$ . The average rate of change from n = 1 to n = 3 is -7.5.

Step-by-step explanation:

From the given graph it is clear that the sequence is a GP because the all terms are half of their previous terms.

Here, $a_2=10,a_3=5,a_4=2.5,a_5=1.25$

$r=\frac{a_3}{a_2}=\frac{5}{10}=\frac{1}{2}$

The common ratio of GP is 1/2.

$r=\frac{a_2}{a_1}$

$\frac{1}{2}=\frac{10}{a_1}$

$a_1=20$

The first term of the sequence is 20.

The formula for sequence is

$a_n=a(r)^{n-1}$

Where a is first term and r is common difference.

The required sequence is

$a_n=20(\frac{1}{2})^{n-1}$

The formula for rate of change is

$m=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

The average rate of change from n = 1 to n = 3 is

$m=\frac{f(3)-f(1)}{3-1}$

$m=\frac{5-20}{3-1}$

$m=\frac{-15}{2}=-7.5$

Therefore the required sequence is $a_n=20(\frac{1}{2})^{n-1}$ . The average rate of change from n = 1 to n = 3 is -7.5.

8 0

3 years ago

Read 2 more answers

william won 50 tickets at the arcade. he redeemed 30 tickets for a prize and gave 5 tickets to ketelyn. which expression can wil

mariarad [96]

Hello!

If William originally won 50 tickets, redeemed 30, and gave 5 to his friend, the expression that represents how many tickets he has left is:

50-30-5

If you are looking for how many tickets William has left, you simply need to solve the expression:
50-30-5 = 15

I hope this helps you! Have a great day!

5 0

3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago