Victor was asked to simplify this expression into exponential form: 8^0 + 12^0.

Find the value of x

Maurinko [17]

Answer:

x=2

Step-by-step explanation:

3 0

2 years ago

Consider the following definitions for sets of characters: Digits = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } Letters = { a, b, c, d, e,

Darya [45]

Answer:

a) Number of passwords: 2763633600

b) Number of passwords: 2487270240

Step-by-step explanation:

We have:

10 digits,

26 letters, and

4 special characters.

We have to compute the number of passwords that satisfy the following conditions:

a) Strings of length 6. Characters can be special characters, digits, or letters, with no repeated characters.

We can use the three types of characters, as long as they do not repeat.

We have a total of 10+26+4=40 characters.

This means we have 40 options for the first place, (40 - 1) for the second place (as the first place already use one of the options available), and so on up to 6 places.

We can write this as:

$C=40*39*38*37*36*35=40!/34!=2763633600$

b. Strings of length 6. Characters can be special characters, digits, or letters, with no repeated characters. The first character can not be a special character.

In this case we add a restriction: the first character can not be a special character.

Then, for the first place we have only 40-4=36 options (digits and letters).

For the second place we can use the 39 options available (digits, letters and special characters).

Then we have:

$C=36*(39*38*37*36*35)=36*(39!/34!)=2487270240$

6 0

3 years ago

Read 2 more answers

The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

cricket20 [7]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

3 0

3 years ago

The product of two consecutive even numbers is 12 more than the square of the smaller number.

vivado [14]

Answer:

6 and 8

Step-by-step explanation:

Let's start off by thinking at what the question is asking us.

The product (multiplication) of two even numbers (x)(x+2) is which means =,

12 more (12) than the square of the smaller number (x).

So put that all together and you get (x)(x+2)=12+x^2=

Distribute the x, (x)(x+2)=

x^2+2x=x^2+12 Subtract x^2 from both sides.

2x=12

Divide both sides by 2 and you get x=6

So the two numbers are 6,8

6(8)=12+6^2

48=48

6 0

3 years ago

What is 5 x 5 I already know the answer but I know people are trying to build up so here you go! Have a fantastic day!

Sever21 [200]

Answer:

25

Extra Info:

Thank you for this !! But for anyone who seriosuy doesn’t know: If you do 5+5, five times, its the same as 5x5 !!

Another Example:

3x3=9

3+3+3=9

3+3=6 6+3=9

7 0

3 years ago