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Yanka [14]
2 years ago
10

7) Halogen lightbulbs last an average of 4,500 hours. If a particular

Mathematics
1 answer:
castortr0y [4]2 years ago
7 0

Answer:

450 days

Step-by-step explanation:

4500/10 = 450 days

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ILL GIVE BRAINLIEST! HELP!
KonstantinChe [14]

Answer:

is B))))

Step-by-step explanation:

brainliest pls

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7 0
2 years ago
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What is the median of the following data set ? {3,4,2,8,5}
Varvara68 [4.7K]

<u>4</u>

<u></u>

Explanation:

3, 4, 2, 8, 5

<u>arrange in ascending order</u>

2, 3, 4, 5, 8

<u>Identify the integer at the middle</u>

median: 4

6 0
2 years ago
Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?
ivanzaharov [21]
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula. 
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C. 
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space. 
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.


8 0
3 years ago
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
GrogVix [38]

Answer:

  1. D. f^-1(x) = log2(x -6)
  2. 4x^2 -3 . . . . x ≤ 0
  3. B. √(x^5) -3√(x^3) -18√x

Step-by-step explanation:

1. When you replace f(x) by x and x by y, you have

... x = 2^y + 6

The first thing you do is subtract 6; then you take the base-2 logarithm:

... (x -6) = 2^y

... log2(x -6) = y = f^-1(x)

You know that to get the y-term by itself, you must subtract 6. Anything else you do will operate on (x-6). Only answer choice D has that sort of construction.

2. When you swap x and y and solve for y, you have ...

... x = -1/2√(y+3)

... -2x = √(y +3) . . . . . . multiply by -2

... (-2x)^2 = y +3 . . . . . square

... 4x^2 - 3 = y = f^-1(x) . . . . subtract 3

The range of f(x) is (-∞, 0], so that is the domain of f^-1(x). That is, f^-1(x) is defined for x ≤ 0.

3. The product of the two functions is ...

... (x -6)(√x)(x +3) = (√x)(x^2 -3x -18)

<em>Every term</em> will have a factor of √x, and the coefficients will be 1, -3, -18. Only selection B matches those conditions.

7 0
3 years ago
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Larry looked at the clock. It was 9:45 p.m. The bus for his class trip leaves at 8:30 am. How many hours and minutes are there u
Mademuasel [1]

Answer:

There are 10 hours and 45 minutes until the bus leaves.

7 0
3 years ago
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