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Arlecino [84]
3 years ago
6

A fair coin is tossed 5 times. what is the probability of no​ heads?

Mathematics
1 answer:
nalin [4]3 years ago
8 0
A tree diagram can be drawn. Since we are not concerned with any result of a head on the coin, we can ignore the branches with heads. To find the probability of the results along the branches, we multiply them, giving rise to the answer 1/32.

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Give the value of the digit 8 in each number.
Anna71 [15]

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construct a two column proof proving SM RN, given RM SN and ∠MRS ∠NSR. Submit the entire proof to your instructor.
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3 years ago
Solve the system by using a matrix equation (Picture provided)
Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]

A= \left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]

X= \left[\begin{array}{c}x\\y\end{array}\right]

B= \left[\begin{array}{c}4\\2\end{array}\right]

i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (7*-6) - (-4*10)

                                    = (-42)-(-40)

                                    = (-42) + 40 = -2

so, |A| = -2

Adj A=  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]

A⁻¹ =  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]/ -2

A⁻¹ =  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]

X= A⁻¹ B

X=  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]

X= \left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]

X= \left[\begin{array}{c}12-4\\20-7\end{array}\right]

X= \left[\begin{array}{c}8\\13\end{array}\right]

x= 8, y= 13

solution set= (8,13).

Option b is correct.

3 0
3 years ago
Use the picture provided to figure out the combination.
AlladinOne [14]

Answer:

4223?

Step-by-step explanation:

i need morw context

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