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Anna11 [10]
3 years ago
15

How to solve (-9)-16 subtracting integers

Mathematics
1 answer:
Ugo [173]3 years ago
5 0
-9-16 = 0-9-16 = 0-25 = -25
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Consider all length-13 strings of all uppercase letters. Letters may be repeated.
serg [7]

a. Your answer is correct, 26 choices for each of 13 positions, so \boxed{26^{13}} total possible strings.

b. You have the right idea, but your method only counts one type of permutation, like

C H A R I T Y _ _ _ _ _ _

but doesn't account for other arrangements like

_ _ _ C H A R I T Y _ _ _

or

_ _ _ _ _ C H A R I T Y _

Treating CHARITY as one letter, we're then considering strings of length 7 (6 open slots plus this string), which we can arrange in 7! different ways. So the total number of such strings is \boxed{7!26^6}.

c. This one is a bit more involved. I would go about it by counting the number of strings containing CHARITY but not HORSES, HORSES but not CHARITY, and both CHARITY and HORSES.

  • CHARITY but not HORSES

As we know from part (a), there are 7!26^6 strings containing CHARITY, but the string HORSES can be found whenever there are 6 open slots to either side of CHARITY, i.e. in strings of either form

C H A R I T Y _ _ _ _ _ _

or

_ _ _ _ _ _ C H A R I T Y

Then there are 2 strings that we want to remove from the count, giving 7!26^6-2 such strings.

  • HORSES but not CHARITY

Reasoning as we did in part (b) suggests that there are 8!26^7 possible strings containing HORSES, and reasoning as we did in the previous case suggests only 2 of these contain CHARITY, giving a total of 8!26^7-2 such strings.

  • CHARITY and HORSES

There are 2 such strings,

C H A R I T Y H O R S E S

H O R S E S C H A R I T Y

Then by the inclusion-exclusion principle, the number of strings containing either CHARITY or HORSES is (7!26^6-2)+(8!26^7-2)-2=7!26^6209-6.

Finally, the number of strings containing neither CHARITY nor HORSES is complementary to the number of strings containing either of them, so the total is 26^{13}-(7!26^6209-6)=\boxed{26^{13}-7!26^6209+6}.

5 0
3 years ago
Alonzo's game started at 9:20 AM.
julsineya [31]

Answer:

11:35 AM

Step-by-step explanation:

Lets start with the hours:

9+2=11

Now the minutes:

15+20=35

Now put them together:

11:35 AM

5 0
3 years ago
Read 2 more answers
The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

y = 2-x \,\,\,\,\,,  1 \leq x \leq 2.

If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

7 0
4 years ago
Simplify: (picture)<br> and please show how you got to the answer, thanks!
Troyanec [42]
(16 a^{3}  b^{4} )^ \frac{3}{4}

Expand,

16^{ \frac{3}{4}  }  a^{3* \frac{3}{4} } b^{4* \frac{3}{4 } }
= 8 a^{ \frac{9}{4} }  b^{3}

So the third choice would be your answer!

Hope this helps!
6 0
3 years ago
Determine whether the following individual event are independent or dependent. Then find the probability of the combined event.
FromTheMoon [43]

Answer:

dependent and 1.26

Step-by-step explanation:

These two individual events are dependent on each other as first they draw it and then instant they eat two red candy pieces

Now the probability of the combined event is as follows

P(Probability of combined event)  is

= P(Event 1) \times P \frac{Event 2}{Event 1}

= \frac{55}{49} \times \frac{54}{48}

= 1.122 \times 1.125

= 1.26

We simply applied the above formula so that we can get the dependency or independency plus the probability of the combined event

                       

5 0
3 years ago
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