How may permutations of the word “spell” are there?

Mathematics

2 answers:

vovangra [49]3 years ago

8 0

I believe that there are 60 permutations of the word spell.

xeze [42]3 years ago

4 0

See, to make a 5-letter word , 5 places are to be filled ___ ___ ___ ___ ___ 1st 2nd 3rd 4th 5th the first place can be filled in 5 ways (s,p,e,l,l) 2nd place in 4 ways. 3rd place in 3 ways. 4th place in 2 ways. 5th place in 1 way. now , ther are 2L's. hence total no. of words that can be formed=(5*4*3*2*1) / 2*1

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jasenka [17]

Answer:

The mean is 42.6

Step-by-step explanation:

$6+90+16+80+15 + 70 + 4 + 60 = 341 \\ there \: are \: 8 \: numbers \: in \: total \: so \: you \: divide \: the \: sum \\ by \: 8. \\ 341 \div 8 = 42.625 \\ mean≈42.6$

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3 years ago

Use the value of the first integral I to evaluate the two given integrals. I = integral^3_0 (x^3 - 4x)dx = 9/4 A. integral^3_0 (

Troyanec [42]

Answer:

A) -9/2

B) 9/4

C) -9/2, same as A)

Step-by-step explanation:

We are given that $I=\int_0^3 x^3-4x dx=9/4$ . We use the properties of integrals to write the new integrals in terms of I.

A) $\int_0^3 8x-2x^3 dx=\int_0^3 -2(x^3-4x) dx=-2\int_0^3 x^3-4x dx=-2I=-9/2$ . We have used that ∫cf dx=c∫f dx.

B) $\int_3^0 4x - x^3 dx=-\int_0^3 (4x-x^3) dx=\int_0^3-(4x-x^3) dx=\int_0^3 x^3-4x dx=I=9/4$ . Here we used that reversing the limits of integration changes the sign of the integral.