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erica [24]
3 years ago
15

A book which cost $300.00 was sold For $240.00. What was the loss percentage

Mathematics
1 answer:
Svetach [21]3 years ago
3 0
Loss % = loss/C.P.*100
= 60/300*100
= 60/3
= 20% loss
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How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird
NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}

Then

Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\

Note that since

F \cap R=\varnothing, Perm(F)\cap Perm(R)\ne \varnothing

But since

B \cap R \ne \varnothing, Perm(B)\cap Perm(R)= \varnothing

and

B \cap F \ne \varnothing , Perm(B)\cap Perm(F)= \varnothing

Since

|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\

where |Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}

and

|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}

What we are looking for is the number of permutations of the 26 letters of the alphabet that do  not contain the strings fish, rat or bird, or

|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0
2 years ago
A prism is filled with 70 cubes with a side length of 1 2 unit. What is the volume of the prism in cubic units? Enter answer as
galina1969 [7]

Answer:

The volume of the cube is;

8\dfrac{3}{4} \ cube \, units

Step-by-step explanation:

The parameters of the prism are;

The number of cubes used to fill the prism, n = 70 cubes

The side length of each cube, s = 1/2 unit

The volume of the cube, V = n × s³

∴ Therefore, the volume of the cube, V = 70 × (1/2)³ = 8\dfrac{3}{4} \ cube \, units = 8.75 cube unit

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V125BC [204]

Answer:

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Step-by-step explanation:

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3 years ago
I would really appreciate if someone would help me i don’t understand
alekssr [168]

Answer:

D. 48

Step-by-step explanation:

In the experiment, 16/40 landed exactly on heads. If you do 16/40 * 3/3 = 48/120.

This means that you can estimate 48 of the next 120 trials have 2 coins showing heads.

6 0
2 years ago
X+y=12<br> 3x = 2y + 6<br> system of elimination
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Answer:

(x,y)=(6,6)

Step-by-step explanation:

there you go

3 0
3 years ago
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