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3 years ago

The sum of the first n odd numbers is n^2. What is the sum of the odd numbers from 1 through 31?

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

6 0

Answer:

256

Step-by-step explanation:

sum of the first n odd number is n^2

if we want to find the sum of of 1 thru 31

to find out how many numbers that is

(1+31)/2 = 32/2 =16

there are 16 numbers

n^2 =16^2 =256

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3 years ago

Solve the following problem then write your answer before the number.

Triss [41]

Answer:

1) Rizza's average speed is 27 km/h.

2) The distance between Barretto and Pampanga is 294 km.

3) Aaron's journey was 3.5 hours long.

Step-by-step explanation:

1) Time taken to cove the distance = 20 minutes

60 minutes = 1 hour

So that, 20 minutes = $\frac{20}{60}$ hours

= $\frac{1}{3}$ hours

Distance covered = 9 km

But,

speed = $\frac{distance covered}{time taken}$

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the car's average speed = 84 km/h

Distance between Barretto and Pampanga = average speed x time taken

= 84 km/h x 3.5 h

= 294 km

The distance between Baretto and Pampanga is 294 km.

3) The uniform speed implies that Aaron did not accelerate during his journey.

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6 0

3 years ago

E + 1 13/16 = 2 5/16

Burka [1]

e + 1 13/16 = 2 5/16

subtract 1 13/16 from each side

e = 2 5/16 - 1 13/16

borrow from the 2

e = 1 16/16 + 5 /16 - 1 13/16

e = 1 21/16-1 13/16

e = 1 8 /16

e = 1 1/2

7 0

3 years ago

Suppose we want to choose 4 objects, without replacement, from 16 distinct objects (a) How many ways can this be done, if the or

lys-0071 [83]

Answer:

a) 1820 ways

b) 43680 ways

Step-by-step explanation:

When the order of the choices is relevant we use the permutation formula:

$P_{n,x}$ is the number of different permutations of x objects from a set of n elements, given by the following formula.

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

When the order of choices is not relevant we use the combination formula:

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

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

In this problem, we have that:

$x = 4, n = 16$

(a) How many ways can this be done, if the order of the choices is not relevant?

$C_{16,4} = \frac{16!}{4!(12)!} = 1820$

(b) How many ways can this be done, if the order of the choices is relevant?

$P_{16,4} = \frac{16!}{(12)!} = 43680$

4 0

3 years ago