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4 years ago

You flip a coin twice. find the probability that at least one of the tosses showed heads.

2 answers:

5 0

1/2, 50%,
In one coin, there is one head and 1 tail
In 2 flips, there are 2 heads and 2 tails
So, all together, there are 4 different possibilities, and 2 are heads
hope this helps!

alexandr402 [8]4 years ago

3 0

Answer:

The probability is 0.75[

Step-by-step explanation:

If we flip a coin twice, we have 4 possibilities, those possibilities are:

TT: we get tails the two times.
TH: On the first flip we get Tails and on the second flip we get Heads.
HT: On the first flip we get Heads and on the second flip we get Tails.
HH: we get Heads the two times.

From these 4 options, 3 show heads at least one of the tosses. These 3 options are TH, HT and HH.

Taking into account that all the possibilities have the same chance, the probability is calculated as the division between 3 and 4 as:

tex]\frac{3}{4} =0.75[/tex]

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Triangle JKL is transformed using the rule . Point K of the pre-image is at (4, 7). What are the coordinates of point K" on the

Zanzabum

Answer:

Option C.

Step-by-step explanation:

Consider the complete question is "Triangle JKL is transformed using the rule $R(O,90^\circ)\circ r_{y=-x}(x,y)$ . Point K of the pre-image is at (4, 7). What are the coordinates of point K'' on the final image?".

According to $r_{y=-x}(x,y)$ , we get

$(x,y)\rightarrow (-y,-x)$

According to $R(O,90^\circ)$ , we get

$(x,y)\rightarrow (-y,x)$

The rule $R(O,90^\circ)\circ r_{y=-x}(x,y)$ means first reflection and then rotation.

We need to find the image of K(4,7).

After reflection, we get

$K(4,7)\rightarrow K'(-7,-4)$

After rotation, we get

$K'(-7,-4)\rightarrow K''(4,-7)$

Therefore, the correct option is C.

8 0

3 years ago

Read 2 more answers

If you were to deposit $1,000 into an account that paid 10 percent interest compounded semiannually, how much money would you ha

dexar [7]

Semiannual means there will be two total periods in a year. Divide ten percent by two to get the interest rate per period. Then multiple by 1+i twice

6 0

3 years ago

Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3

nlexa [21]

Answer:

$(x,y) = (5,-6)$

Step-by-step explanation:

$\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\$

$~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)$

$\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\$

$\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5$

$y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6$