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melamori03 [73]
3 years ago
8

A light fixture at a traffic intersection operates on exactly 500 watts of power to light the bulbs in the fixture. Any combinat

ion of 150-watt, 100-watt, 75-watt, 60- watt or 50-watt bulbs can be used but the total number of watts must be 500. More than one of the same watt can be used. How many combinations could be used if at least one bulb must be 60 watt?
Mathematics
1 answer:
andrey2020 [161]3 years ago
7 0

Answer: I found 5 combinations, key thing is that you need to have 5

60w bulbs

Step-by-step explanation: if we need to have at least one 60 W bulb, we notice

Five 60 W bulbs make 300W. Then we get missing 200 W with following combinations: 100 W + 100 W , 75 + 75 + 50 , 150 + 50 , 100+ 50 + 50

Or 4 times 50 W.

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Determine wether 16 is the perfect square show proof
enot [183]

Answer:

Therefore, the square root of 16 is an integer, and as a consequence 16 is a perfect square.

As a consequence, 4 is the square root of 16.

Step-by-step explanation:

A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 16 is 4.

Therefore, the square root of 16 is an integer, and as a consequence 16 is a perfect square.

As a consequence, 4 is the square root of 16.

5 0
4 years ago
What is the value of k^16 - 16^k, if x = 13^2 * 15^4 * 17^6 * 19^8 and is a multiple of 16^k, where k is a whole number?
Andrei [34K]

Answer:

I tried but I kept getting weird numbers

Step-by-step explanation:

5 0
3 years ago
Samantha invests $600 at a rate of 2% per year simple interest.
bulgar [2K]

Answer:

$96

Step-by-step explanation:

2% times 8 years equals 16, then multiply 0.16 by $600 and you get $96.

8 0
3 years ago
You know how to play 21 songs on your guitar. (a) If you want to choose 8 of the songs to make a setlist to play for a small gat
UNO [17]

Answer:

<em>(a) 8,204,716,800</em>

<em>(b) 5,985</em>

Step-by-step explanation:

<u>Combinations and Permutations</u>

Combinatorics is the part of the discrete mathematics that studies the enumeration of groups or sorting of a determined number of elements.  The concept of combinations is tied to the different forms to group elements where the order of their arrangements is not important or does not differentiate from the very same set of elements picked in a different order.

The concept of combinations is tied to the differents forms to group elements where the order of their arrangements is not important or does not differentiate from the very same set of elements picked in different order.

On the other hand, permutations or variations are sets selected in a specific order and another set with the same element but in different order is considered a different set.

If we have n elements available to pick from in sets of m elements each, there can be C(n,m) different combinations, and it's given by

\displaystyle C(n,m)=\frac{n!}{m!(n-m)!}

Similarly the number of permutations is given by

\displaystyle P(n,m)=\frac{n!}{(n-m)!}

(a) I have n=21 songs to pick from and I want to choose m=8 of them where the order matters, so it's a permutation:

\displaystyle P(21,8)=\frac{21!}{(13)!}=\frac{51,090,942,171,709,440,000}{6,227,020,800}=8,204,716,800

I can make more than 8 billion different setlists

(b) To choose m=4 songs from n=21 songs where the order does not matter, we compute the combination

\displaystyle C(21,4)=\frac{21!}{4!(17)!}=\frac{21\cdot 20\cdot 19\cdot 18\cdot 17!}{4\cdot 3\cdot 2\cdot 1(17)!}

\displaystyle C(21,4)=\frac{143,640}{24}=5,985

I can make almost 6,000 sets of 4 songs

5 0
3 years ago
Simplify |a+3| + |a−3| , if −3
Rashid [163]

Answer:

6

Step-by-step explanation:

I'm not sure if you mean whether a = -3, but if you are to plug in -3 for a you should get:

|-3+3|+|-3-3|\\=|0|+|-6|\\=0+6\\=6

5 0
4 years ago
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