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Temka [501]
3 years ago
8

Help please 5 points

Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

the answer to this question is b

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Solve the system of equations​
34kurt

Answer:

y=17/2 and x=15/8

Step-by-step explanation:

By subsitution, we get 4x+4x+1=16, so 8x=15. Thus, x=15/8. Next, we plug x in to find y. y=4*15/8 +1, or 15/2 +1 = 17/2

5 0
3 years ago
Read 2 more answers
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
3 years ago
6. Which of the following fractions is equivalent to
hichkok12 [17]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
Need help...with this graph
Lady_Fox [76]

Answer:

The graph with coordinates x6,y8

Step-by-step explanation:

7 0
3 years ago
Zoey made 6 3/4 cups of fruit salad for a picnic. At the picnic, they ate 1/3 of the fruit salad. How much of the fruit salad di
Aleks04 [339]

Answer:

9/4 cups

Step-by-step explanation:

6 3/4 = 27/4

27/4 * 1/3 = 27/12

27/12 = 9/4

4 0
3 years ago
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