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kvv77 [185]
3 years ago
7

Slop intercept form of: y-5=2(x+2)

Mathematics
1 answer:
saw5 [17]3 years ago
6 0

Answer: y= 2x+9

have a good day!

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I'm having g a hard time with this, anyone please care to solve it for me? I would highly appreciate it.​
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43 46 34 98 78 67

Step-by-step explanation:

8 0
3 years ago
Find the length of XY.<br> Imported Asset <br><br> 4<br><br> 8<br><br> 16<br><br> 11.3
kompoz [17]

Answer: 8 (The second option).


Step-by-step explanation:

1. To solve this exercise you draw a line to a point O, this lines will bisect the angle of Z∠60°. Then will obtain two into two right triangles with two angles of 30° at the point Z.

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7 0
3 years ago
3x + (-9x) + 12 + 8
ohaa [14]
(3x−9x)+(12+8)
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8 0
3 years ago
Read 2 more answers
Gummi bears come in twelve flavors, and you have one of each flavor. Suppose you split your gummi bears among three people (Adam
aliina [53]

Step-by-step explanation:

From the given gummi bear, the chance that Adam is selected for any draw is 1/3 as well as the chance he is not selected at any draw is 2/3.

a). The probability of Adam getting exactly three gummi bears = P(Adam gets selected at 3 draws and not selected at the remaining 9 draws)

= $ (\frac{1}{3})^3 (\frac{2}{3})^9 = \frac{2^9}{3^{12}} $

Now, the 3 draws where Adam gets selected can be any 3 out of 12 draws in $ {\overset{12}C}_3 $ = 220

Thus, probability of Adam getting three gummi bears = $ 220 \times \frac{2^9}{3^{12}} $

                                                                                                = 0.21186

b). Probability that Adam will get the three gummi bears given each person will received at the most 1 gummi bear

= P(of the remaining 9 draws after assigning one gummi bear to each one, Adam gets selected at 2 draws and not selected at 7 draws) = $ {\overset{9}C}_2 (\frac{1}{3})^2 (\frac{2}{3})^7 $

                                                                                               = 0.23411

c). Let X = Number of the gummi bears which Adam will get. Then, X = number of draws out of 12 draws Adam gets selected and X ~ B(12, 1/3). So,  Adam will get gummi bears= mean of B(12, 1/3) = 12 x (1/3) = 4

d). Let Y = Number of the gummi bears that Adam will get, given each person will received at the most 1 gummi bear Then, Y = number of draws Adam gets selected in the remaining 9 draws and Y ~ B(9, 1/3). So, the expected number of bears that Adam gets given each person received at least 1 gummi bear

= 1 + mean of B(9, 1/3) = 4

e). When every one gets atleast one gummi bear, the new sample size will be 9 and so we can say that there is a reduction in variance.

8 0
3 years ago
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