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Angelina_Jolie [31]
3 years ago
8

A laundry basket contains 18 blue socks and 24 black socks. What is the probability of randomly picking 2 black socks, without r

eplacement, from the basket?
Mathematics
1 answer:
lisov135 [29]3 years ago
6 0
P(black socks): 24/42 or 12/21
P(black socks without replacing): 23/41. As a result, the probability of randomly picking 2 black socks, without replacement, from the basket is 12/21×23/41=276/861 or 32%. Hope it help!
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Please helpppppppppp<br> include step by step
balandron [24]

Answer:

507

Step-by-step explanation:

As they said, I will be using 3 as pi.

3 (pi) * (\frac{26}{2})^2 (radius squared)

Without annotations: 3*\frac{26}{2}^2

3*\frac{26}{2}^2

3*13^2; 13*13=169

3*169

507



5 0
2 years ago
If R is the midpoint of QS, QR= 8x-51 and RS = 3x-6, find QS.
sashaice [31]

QS would be 42. So the way i did it was I:

8x-51=3x-6

x=9

Plugged into both equations of QR and RS to then get the answer of 42

7 0
3 years ago
Read 2 more answers
The sum of two numbers is 42 and the difference is 16
mamaluj [8]
The answer for this one is 29 and 13. Hope it help!
4 0
3 years ago
Write an exponential function y = abx for a graph that includes (2, 24) and (3, 48).
labwork [276]
I think you mean y=ab^x.
First, plug in the values of x=2 and y=24 for the first equation, and x=3 and y=48 for the second one:

24 = ab^2
48 = ab^3

divide equation 2 by equation 1:
2 = b^1

or, more simply:
2 = b

now plug b into one of the equations to get "a":

24 = a*2^2
24 = a*4
6 = a

So, the answer is y = 6(2)^x
(I think you need to specify the power of x in your answers)
7 0
3 years ago
Read 2 more answers
Please help me i cant get this i. honestly just need someone to give the answers for this lesson
Likurg_2 [28]
Area of the triangle = (1/2)*base* height = (1/2)*3*6 = 9 sm²
We have 4 triangles like this one ,
so area of 4 triangles  = 9*4 = 36 cm².
Area of the square = 3*3 = 9 cm²

Surface area here means area all triangles  + area of the square =36 +9 =
= 45 cm ²

Answer is 45 cm².

7 0
3 years ago
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