Ask question

Ask question

All categories

3 years ago

9

using a list or tree diagram, show all the possible outcomes for tossing three coins one time, using the letter H when a coin fa

ces “heads up” and the letter T when is faces “tails up”

1 answer:

tresset_1 [31]3 years ago

4 0

I just did this in math class so
t h
h t h t

h t h t ht ht

You might be interested in

15/4 in to equal mixed number

Temka [501]

9514 1404 393

Answer:

3 3/4

Step-by-step explanation:

$\dfrac{15}{4}=\dfrac{12+3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\\\\=3+\dfrac{3}{4}=\boxed{3\dfrac{3}{4}}$

__

The integer part of the mixed number is the quotient when the numerator is divided by the denominator. The remainder from that division is the numerator of the fraction.

15/4 = 15 ÷ 4 = 3 r 3 = 3 3/4

5 0

2 years ago

Add.<br> (8y + 6) + (4y + 5)<br> Submit

Inga [223]

Answer:

12y + 11

Step-by-step explanation:

To add them you would add like terms. That means you would add those with y together and those without together.

8y + 4y + 6 + 5 = 12y + 11

If you want to solve for y, subtract 11 from both sides then divide both sides by 12.

6 0

3 years ago

Read 2 more answers

HELP AHHH<br> PLEASE AND THANK YOU

aleksley [76]

Is it D? if not i’m so sorry

4 0

2 years ago

Read 2 more answers

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

What is 4/9 in a decimal form

Delicious77 [7]

0.44...........is the answer

8 0

3 years ago

Read 2 more answers