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Luba_88 [7]
3 years ago
7

if birthdays are equally likely to fall on any day, what is the probability that a person chosen a random has a birthday in janu

ary?
Mathematics
2 answers:
deff fn [24]3 years ago
6 0
Total days in a year = 365
days in January = 31

Probability of having birthday in January = 31/365 = 0.0849
Elan Coil [88]3 years ago
5 0
The probability a birthday that falls in January is: 

<span>≈ 7.734% </span>
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7. y-2 = (x+3)<br> 4<br> 2<br> X<br> - TO<br> 12<br> 4
drek231 [11]

Answer:

y = -24 + -2x + 2x2

Step-by-step explanation:

Simplifying:

y = 2(x + 3)(x + -4)

reorder the terms:

y = 2(3 + x)(x + -4)

multiply (3 + x) * (-4 + x)

y = 2(3(-4 + x) + x(-4 + x))

y = 2((-4 * 3 + x * 3) + x(-4 + x))

y = 2((-12 + 3x) + x(-4 + x))

y = 2(-12 + 3x + (-4 * x + x * x))

y = 2(-12 + 3x + (-4x + x2))

Combine like terms  3x + -4x = -1x

y = 2(-12 + -1x + x2)

y = (-12 * 2 + -1x * 2 + x2 * 2)

y = (-24 + -2x + 2x2)

Solving:

y = -24 + -2x + 2x2

solving for variable 'y'

Move all terms containing y to the left, all other terms to the right.

Simplifying:

y = -24 + -2x + 2x2

8 0
3 years ago
In a basketball game, a regular basket was worth 2 points and a long-distance basket was worth 3 points. if there were 45 basket
IgorLugansk [536]
Let
 x: number of regular basketball
 y: number of long-distance basket
 We have the following system of equations:
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 x + y = 45
 Solving the system we have
 y = 45-x
 2x + 3 (45-x) = 96
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 -x = 96 -135
 x = 39
 Then,
 y = 45-x
 y = 45-39
 y = 6
 answer
 were made
 regular baskets = 39
 long-distance baskets = 6
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3 years ago
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Lerok [7]

Answer:

Step-by-step explanation:

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3 years ago
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Troyanec [42]
56, because you times 7 by 2 because Greg has 2 then you times 7 by 10 because then you get the answers 14 and 70 and take them away from each other
7 0
3 years ago
Can you guys help me out pls
kozerog [31]
Answer is for sure b
8 0
3 years ago
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