Find the 50th term of this arithmetic sequence: 6, 13, 20, 27, . . . Answer:

Two number have a sum of 71 and a difference of 37

larisa86 [58]

The right answer for the question that is being asked and shown above is that:

Two numbers have a sum of 71 and a difference of 37
x + y = 71
x - y = 37

So in order to get the two numbers, here it is:
x + y = 71
x - y = 37
------------
2x = 108
x = 54
y = 71 - 54
y = 17

So the two numbers are 54 and 17

6 0

3 years ago

Read 2 more answers

32% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

Tems11 [23]

Answer:

a) $P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

b) $P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

c) $P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.32)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

Part b

$P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]$

$P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211$

$P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

Part c

$P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264$

$P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218$

$P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123$

$P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

6 0

3 years ago

A box has colored balls (12 red, 10 green, 6 light blue and 8 yellow). What is the probability of drawing 2 red balls?

ale4655 [162]

Answer:

You will have a 33% chance of drawing a red ball

Step-by-step explanation:

If you do 12+10+8+6, you get 36 which then you would ratio it. 12:36, and you would convert that to a percent, which is 33%

7 0

2 years ago

the table shows the balance of a money market account over time. write a function that represents the balancr y (in dollars) aft

Solnce55 [7]

Answer:

y = 1500 * 1.09^t

Step-by-step explanation:

It should be clear that the rate of change is non-linear, meaning that we can express the function in the following manner

y = a * b^t, where a is the initial value and b is the rate of change and t is how many years.

Well, year 0 is the rate of change so we can say that a = 1500

Next, we divide any two consecutive year balances to find our rate of change:

1635/1500 = 1.09

Therefore, our equation looks like this: y = 1500 * 1.09^t

6 0

3 years ago

In a race in which eleven automobiles are entered and there are no ties, in how many ways can the first three finishers come in?

viva [34]

Answer:

990 ways

Step-by-step explanation:

The total number of automobiles we have is 11.

Now, what this means is that for the first position , we shall be selecting 1 out of 11 automobiles, this can be done in 11 ways( 11C1 = 11!/(11-1)!1! = 11!/10!1! = 11 ways)

For the second position, since we have the first position already, the number of ways we can select the second position is selecting 1 out of available 10 and that can be done in 10 ways(10C1 ways = 10!9!1! = 10 ways)

For the third position, we have 9 automobiles and we want to select 1, this can be done in 9 ways(9C1 ways = 9!/8!1! = 9 ways)

Thus, the total number of ways the first three finishers come in = 11 * 10 * 9 = 990 ways

8 0

3 years ago