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3 years ago

How many different strings can be made from the letters in HAPPINESS, using all the letters?

Mathematics

1 answer:

Rasek [7]3 years ago

5 0

Answer:

20 different strings

Step-by-step explanation:

Let the lines in the alphabets be the strings. So, each alphabet has many strings. Counting them would make them 20 strings.

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11.(02.03)

Marianna [84]

Answer:

step number 1 is incorrect.

Step-by-step explanation:

Here is the solution of this equation [2x-31-1=2].

2x-31-1=2

or 2x-31=2+1

or 2x=2+1+31

or 2x= 34

or x=34/2

or x= 17

the answer of this equation is X=17.

8 0

3 years ago

Anton Louis is a junior partner in an accounting firm. He is single

Sauron [17]

Answer:

Step-by-step explanation:

3 kids

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=y%3Dx%2B3y%3Dx-1" id="TexFormula1" title="y=x+3y=x-1" alt="y=x+3y=x-1" align="absmiddle" class

Dominik [7]

Answer:

y = x + 3y = x - 1

Step-by-step explanation:

Just graph them!

Hope I could help.

5 0

3 years ago

A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one

d1i1m1o1n [39]

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

$P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855$

The probability that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.

8 0

3 years ago

“Find the values of x and y when the smaller triangle here has the given area! “

natita [175]

Answer:

I think it’s 48 cm

Step-by-step explanation:

Area is length x width so 8 x 12 is 96 and since it is twice the slice of the smaller triangle, 96 divided by 2 is 48

3 0

3 years ago