All categories

2 years ago

I need held asp pls help me

Mathematics

1 answer:

aksik [14]2 years ago

5 0

Answer:

18/23=18/23

Step-by-step explanation:

This problem is already in simplest form.

Hope this helped!!!!

You might be interested in

What is<br> 15/100<br> as a percentage?

Lunna [17]

Answer:

15%

Step-by-step explanation:

hope i helped!! please follow ❤❤

7 0

2 years ago

Read 2 more answers

A bag contains 10 tiles with the letters A, B, C, D, E, F, G, H, I, and J. Five tiles are chosen, one at a time, and placed in a

lora16 [44]

I assume in this item, we are to find at which step is the mistake done for the calculation of the unknown probability.

For the possible number of arrangement of letter, n(S), the basic principles of counting should be used.
= 10 x 9 x 8 x 7 x 6 = 30,240

This is similar as to what was done in Meghan's work.

For the five tiles to spell out FACED, there is only one (1) possibility.

Therefore, the probability should be equal to 1/30,240 instead of the 1/252 which was presented in the steps above.

8 0

2 years ago

wel

A is the corect answer they congruent.

8 0

3 years ago

Read 2 more answers

Frank, mary and seth shared some sweets in the ratio 4;5;7

Rashid [163]

Answer:

total = 96 sweets

Step-by-step explanation:

sweets shared in the ratio 4 : 5 : 7 = 4x : 5x : 7x ( x is a multiplier )

Seth got 18 more sweets than Frank , that is

7x = 4x + 18 ( subtract 4x from both sides )

3x = 18 ( divide both sides by 3 )

x = 6

Then

total = 4x + 5x + 7x = 16x = 16 × 6 = 96 sweets

4 0

2 years ago

Solve the following by using mathematical Induction. For >/ 1

cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

in left part $1^2=1;$

in right part $\dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.$

2. Assume that for all $k$ following equality is true

$1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

3. Prove that for $k+1$ the following equality is true too.

$1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$

Consider left part:

$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$

Consider right part:

$\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)$

We get the same left and right parts, so the equality is true for $k+1.$

By mathematical induction, this equality is true for all n.

3 0

3 years ago