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3 years ago

Can someone help me please

Mathematics

1 answer:

disa [49]3 years ago

3 0

Answer:

We're told that the chance of landing heads is 2/7, which means that the chance of landing a tail is 5/7.

The chance of rolling each combination then is:

h, h: (2/7) * (2/7) = 4/49

h, t: (2/7) * (5/7) = 10/49

t, h: (5/7) * (2/7) = 10/49

t, t: (5/7) * (5/7) = 25/49

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Factor <img src="https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="-\frac{1}{3}" alt="-\frac{1}{3}" align=

ivanzaharov [21]

10/3 I did the math

4 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago

Hhhhhhhheeeeeeeeeellllllllllpppppppppp

juin [17]

answer: d

explantion :

Factor the numerator and denominator and cancel the common factors.

3 0

2 years ago

7. Which coordinate lies in the solution set for

Nana76 [90]

Answer:

The only point (0,0) lies inside the shaded region and hence it gives a solution for the set of inequalities.

Step-by-step explanation:

See the graph attached to this question.

The solution of the set of inequalities is given by the shaded region on the graph.

Now, the point (0,5) is outside this shaded region, hence it can not be the solution.

The point (3,0) also is outside this shaded region, hence it can not be the solution.

The point (-3,0) also is outside this shaded region, hence it can not be the solution.

Now, the only point (0,0) lies inside the shaded region and hence it gives a solution for the set of inequalities. (Answer)

6 0

2 years ago

Can anyone help with this?

Karo-lina-s [1.5K]

The answer is B. If you take the even root of a positive number, you'll get two solutions, a positive and a negative

4 0

3 years ago