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Ymorist [56]
3 years ago
9

When you flip a biased coin the probability of getting a tail is 0.51.

Mathematics
1 answer:
pickupchik [31]3 years ago
7 0

Answer:

0.49

Step-by-step explanation:

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The question is in the picture. Plz helppp it is due tonight
Mkey [24]

Answer:

Step-by-step explanation:

We're given one equation, we have to find the other equation and solve each for 200, and whichever has the lower x is the winner

for Jayden we are given

(0, 48), (1, 56), (2, 64), (3, 72), (4, 80)

This looks to be a line. The y values are each separated by a common difference

We can use two points to describe the line

m = \frac{y_2-y_1}{x_2-x_1} = \frac{80-48}{4-0} = \frac{32}{4} = 8\\  K = 8m + b\\48 = 8(0) + b\\b = 48\\so\\K = 8m + 48

Now we can set both K = 8m + 48 and K = m^2 + 10 equal to 200

200 = 8m + 48\\152 = 8m \\m = \frac{152}{8} = 19

200 = m^2 + 10\\m^2 = 190\\m = \sqrt{190} = 13.784

Keiko's blog will reach 200 subscribers fastest.

7 0
3 years ago
Please help me now please
Kryger [21]

Answer:

0.4444444

Step-by-step explanation:

2/3 x 2/3 = 0.4444 repeating!

Hope this helps!

7 0
3 years ago
2+x^2=18<br> Help please
Harlamova29_29 [7]

Answer:

2+x^2=18 = 2+4^2=18

Step-by-step explanation:

Because when you have the ( ^2) then that means your going to multiply a number by the same number and then 4 times 4 equals 16 and you still have the +2 therefore you add that on the 16 and 16+2 equals 18.

3 0
3 years ago
Can a number be both rational and irrational? Why or why not?
AURORKA [14]

Answer:

Step-by-step explanation:

If it is irrational, it can never be rational unless it is acted upon by something else like a power.

Pi, no matter what is done to it, can never be rational.  Once a number is declared as irrational, that is what it remains.

6 0
2 years ago
Which of the following are solutions to 2tanx/1-tan^2x=sqrt 3?
Tanya [424]

Answer:

A, D

Step-by-step explanation:

(\frac{2tanx}{1-tan^{2}x}) =\sqrt{3} (1)

Assume that tan x = y. Replace this into the equation (1), we have:

+) \frac{2y}{1-y^{2} } =\sqrt{3}

=> 2y = \sqrt{3}  . (1 -y^{2} ) =- \sqrt{3} .y^{2}  + \sqrt{3}

=> \sqrt{3} .y^{2}  + 2y - \sqrt{3} = 0

=> \sqrt{3} . y^{2} - 1.y + 3.y - \sqrt{3} =0

=> (\sqrt{3} . y.y - 1.y) + (\sqrt{3}.\sqrt{3}  .y - \sqrt{3}) =0

=> y.(\sqrt{3} y - 1) + \sqrt{3} (\sqrt{3} y - 1) = 0

=> (y + \sqrt{3} ).(\sqrt{3} .y -1)=0

=> y + \sqrt{3} = 0 or \sqrt{3} y -1 =0

If y + \sqrt{3} = 0

=> y = - √3

=> tan x = - √3

=> x = \frac{2\pi }{3} + n.\pi with n is an integral

If \sqrt{3} y -1 =0

=> y = 1/√3

=> tan x = 1/√3

=> x = \frac{\pi }{6} + n.\pi n is an integral

So that A, D are answers.

3 0
3 years ago
Read 2 more answers
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