All categories

3 years ago

How many whole numbers are less than n+1

Mathematics

1 answer:

andreev551 [17]3 years ago

6 0

Answer:

$n$ .

Step-by-step explanation:

Assuming that $n \in Z^{+}$ , then all number less than $n+1$ are

$n+1-1=n$

Notice that the "plus one" indicates that the whole number shown by the expression $n+1$ indicates a whole number greater than $n\\$ by one unit. So, we just subtract one whole unit to achive a number less than the given expression.

Therefore, the answer is $n$ .

You might be interested in

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

3 years ago

A video rental store keeps a list of their top 15 movie rentals each week. This week the list includes 6 action, 4 comedies, 3 d

azamat

Answer:

Step-by-step explanation:

Probability is the likelihood or chance that an event will occur.

Probability = Expected outcome of event/Total outcome.

If a video rental store keeps a list of their top 15 movie rentals each week, the total outcome is 15.

If the list for the week includes 6 action, 4 comedies, 3 dramas, and 2 mysteries and the store manager removes a copy of each of the 15 movies from the shelf, then randomly selects 3 of the 15 to show on the display monitors in the store, the probability that she selected 2 comedies and 1 action movie will be calculated as shown;

Probability of selecting 2 comedies = 4/15*4/15 = 16/225 (Note that the expected outcome in this case is 4).

Probability of selecting 1 action movie = 6/15 = 2/5

<em>Hence, the probability that she selected 2 comedies and 1 action movie will be equivalent to 16/225*2/5 = 32/1125</em>

<em>Note that the rented movies will have to be returned hence reason for the replacement. </em>

8 0

3 years ago

Help with 5b please . thank you.

Allushta [10]

Answer:

See explanation

Step-by-step explanation:

We are given f(x)=ln(1+x)-x+(1/2)x^2.

We are first ask to differentiate this.

We will need chain rule for first term and power rule for all three terms.

f'(x)=(1+x)'/(1+x)-(1)+(1/2)×2x

f'(x)=(0+1)/(1+x)-(1)+x

f'(x)=1/(1+x)-(1)+x

We are then ask to prove if x is positive then f is positive.

I'm thinking they want us to use the derivative part in our answer.

Let's look at the critical numbers.

f' is undefined at x=-1 and it also makes f undefined.

Let's see if we can find when expression is 0.

1/(1+x)-(1)+x=0

Find common denominator:

1/(1+x)-(1+x)/(1+x)+x(1+x)/(1+x)=0

(1-1-x+x+x^2)/(1+x)=0

A fraction can only be zero when it's numerator is.

Simplify numerator equal 0:

x^2=0

This happens at x=0.

This means the expression,f, is increasing or decreasing after x=0. Let's found out what's happening there. f'(1)=1/(1+1)-(1)+1=1/2 which means after x=0, f is increasing since f'>0 after x=0.

So we should see increasing values of f when we up the value for x after 0.

Plugging in 0 gives: f(0)=ln(1+0)-0+(1/2)0^2=0.

So any value f, after this x=0, should be higher than 0 since f(0)=0 and f' told us f in increasing after x equals 0.

8 0

2 years ago

the diameter of the circle base of a storage container 18.8 m circumference of the base is approximately 59 m which of these cou

ozzi

Answer:

Estimate the value of pi =3.14

Step-by-step explanation:

The diameter of the circle base of a storage container 18.8m

The diameter of the circle 'd' = 2r

Given data 'd' = 2r = 18.8m

The circumference of the circle = 2πr

Given data of circumference of the circle = 59 m

given 2πr = 59

2r(π) = 59

18.8 π = 59 ( by using 2r =18.8)

π = 59/18.8

π = 3.13829

π = 3.14

<u>Conclusion</u>:-

Estimate the value of pi =3.14

6 0

3 years ago

How do you simplify Radicals?<br>√180v^4

julsineya [31]

Here are the steps required for Simplifying Radicals:

Step 1: Find the prime factorization of the number inside the radical. Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Also factor any variables inside the radical.

Step 2: Determine the index of the radical. The index tells you how many of a kind you need to put together to be able to move that number or variable from inside the radical to outside the radical. For example, if the index is 2 (a square root), then you need two of a kind to move from inside the radical to outside the radical. If the index is 3 (a cube root), then you need three of a kind to move from inside the radical to outside the radical.

Step 3: Move each group of numbers or variables from inside the radical to outside the radical. If there are nor enough numbers or variables to make a group of two, three, or whatever is needed, then leave those numbers or variables inside the radical. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group.

Step 4: Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Multiply all numbers and variables outside the radical together.

Shorter version:

Step 1: Find the prime factorization of the number inside the radical.

Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.

Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and 3’s moved outside the radical.

Step 4: Simplify the expressions both inside and outside the radical by multiplying.

7 0

2 years ago