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eimsori [14]
3 years ago
15

Find the number of 4-digit numbers that contain b at least three odd digits.

Mathematics
1 answer:
melamori03 [73]3 years ago
6 0

Answer:

  3000

Step-by-step explanation:

There are 5 odd digits, so 5^3 = 125 ways to have 3 odd digits.

Given 3 odd digits, the first digit can be even, but not zero, so there are 4 ways to do that.

The second, third, or fourth digits can be even 5 ways, so there are ...

  125(4 +5 +5 +5) = 2375

ways to have 3 odd digits in a 4-digit number.

__

There are 5^4 = 625 ways to have 4 odd digits.

The number of 4-digit numbers with 3 or 4 odd digits is ...

  2375 +625 = 3000

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12 and 3000 as decimals full equation
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Answer:

12=0.12x100

3000=0.3x10000

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2 years ago
What is the square root of 149 rounded to the nearest tenth?
Ad libitum [116K]
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3 years ago
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The sum of four consecutive even integer numbers is 84. find the four numbers.
vodomira [7]

84/4=21

 now take the 2 even numbers below 21 and the 2 even numbers above 21

18 +20 + 22 +24 = 84

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3 0
3 years ago
How many 1/4 cups in 3 1/2 cups
seropon [69]
If you have 3 1/2 cups of something and you want to find how many 1/4 cups there are the you just need to divide 3 1/2 by 1/4
3 \frac{1}{2} ÷ \frac{1}{4}
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hope I helped :)
6 0
3 years ago
Consider the initial value problem y′′+36y=2cos(6t),y(0)=0,y′(0)=0. y″+36y=2cos⁡(6t),y(0)=0,y′(0)=0. Take the Laplace transform
Bingel [31]

Recall the Laplace transform of a second-order derivative,

L(y''(t)) = s^2Y(s)-sy(0)-y'(0)

and the transform of cosine,

L(\cos(at))=\dfrac s{a^2+s^2}

Here, both y(0)=y'(0)=0, so taking the transform of both sides of

y''(t)+36y(t)=2\cos(6t)

gives

s^2Y(s)+36Y(s)=\dfrac{2s}{36+s^2}

\implies Y(s)=\dfrac{2s}{(s^2+36)^2}

4 0
3 years ago
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