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2 years ago

A fair coin is flipped two times. Which diagram shows all of the possible outcomes?

Mathematics

1 answer:

kirza4 [7]2 years ago

6 0

Answer:

Step-by-step explanation:

It is the only one that has both heads and tails equally on all options, instead of heads heads or tails tails. There aren't any fair coins that are like that

hope this helps :)

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A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers

Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

$\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57$

Using normal distribution table, $z_{0.005$ is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

$e= z_{0.005}\frac{\sigma}{\sqrt{n} } = 2.57*\frac{3900}{\sqrt{100} } =1002.3$

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) = (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0

4 years ago

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

5 0

3 years ago

Read 2 more answers

The sales tax in New Jersey is 6%. For a $16 book, what will the total cost be, including sales tax?

Rasek [7]

<span>The sales tax in New Jersey is 6%. For a $16 book, what will the total cost be, including sales tax?
Solutions:
=> 16 dollars is the amount of book
=> 6% is the sales tax which is equals to (6% / 100% = 0.06)
Let’s follow the formula to get the total amount to be paid.
=> 16 dollars * .06 = .96 dollars
Now, let’s add this together:
=> .96 + 16 dollars = 16.96 dollars.</span>

5 0

3 years ago

Read 2 more answers

Does anybody know the answer?

Anon25 [30]

Answer:

0.00000381469

Step-by-step explanation:

there u go

4 0

3 years ago

(A)using geometry vocabulary, describe a sequence of transformations that maps figure P (-1,2)(-1,4) (-4,2) (-4,4) onto figure Q

andrey2020 [161]

Before we proceed on determining the transformation happening on this problem, it's better to see first the location of the figure by drawing it in a cartesian coordinate plane. We have

If we observe the figures and the coordinates of the plot, we can see that there is a difference of 1 on the x coordinates of P and y coordinates of Q. Therefore, the first transformation that we consider here is the movement of figure P by 1 unit to the left. We have

$\begin{gathered} P_1=(-1-1,2_{})=(-2,2) \\ P_2=(-1-1,4)=(-2,4) \\ P_3=(-4-1,2)=(-5,2) \\ P_4=(-4-1,4)=(-5,4) \end{gathered}$

This transformation changes the location of figure P into

The next transformation will be the rotation of the red dotted figure on the figure above by 90 degrees counterclockwise. With this transformation, the coordinates will transform as

$P_{ccw,90}=(-y,x)$

Hence, for the rotation, we have the new coordinates.

$\begin{gathered} P_1^{\prime}=(-2,-2) \\ P_2^{\prime}=(-4,-2) \\ P_3^{\prime}=(-2,-5) \\ P_4^{\prime}=(-4,-5) \end{gathered}$

The transformed image, which is represented as NMPO, will now be at

For the last transformation, we will be reflecting the figure NMPO over the <em>y</em> axis. This changes the coordinates as

$P_{\text{rotation,y}-\text{axis}}=(-x,y)$

We now have the new coordinates:

$\begin{gathered} P^{\doubleprime}_1=(2,-2)=Q_1_{}_{} \\ P_2^{\doubleprime}=(4,-2)=Q_3 \\ P_3^{\doubleprime}=(2,-5)=Q_2 \\ P_4^{\doubleprime}_{}=(4,-5)=Q_4_{} \end{gathered}$

As you can see, they have the same coordinates as figure Q.

The mapping rules for the sequence described above are as follows:

First transformation (moving one unit to the left (x-1,y))

$\begin{gathered} P_1(-1,2)\rightarrow P_1(-1-1,2)\rightarrow P_1(-2,2) \\ P_2(-1,4)\rightarrow P_1(-1-1,4)\rightarrow P_2(-2,4) \\ P_3(-4,2)\rightarrow P_1(-4-1,2)\rightarrow P_3(-5,2) \\ P_4(-4,4)\rightarrow P_1(-4-1,4)\rightarrow P_4(-5,4) \end{gathered}$

Second transformation (rotation counter clockwise (-y,x))

$\begin{gathered} P_1(-2,2)\rightarrow P^{\prime}_1(-2,-2)_{} \\ P_2(-2,4)\rightarrow P^{\prime}_2(-4,-2) \\ P_3(-5,2)\rightarrow P^{\prime}_3(-2,-5)_{} \\ P_4(-5,4)\rightarrow P^{\prime}_4(-4,-5)_{} \end{gathered}$

Third Transformation (reflection over y-axis (-x,y))

$\begin{gathered} P^{\prime}_1(-2,-2)\rightarrow P^{\doubleprime}_1(-(-2),-2)\rightarrow P^{\doubleprime}_1=(2,-2)=Q_1 \\ P^{\prime}_2(-4,-2)\rightarrow P^{\doubleprime}_1(-(-4),-2)\rightarrow P^{\doubleprime}_1=(4,-2)=Q_3 \\ P^{\prime}_3(-2,-5)\rightarrow P^{\doubleprime}_1(-(-2),-5)\rightarrow P^{\doubleprime}_1=(2,-5)=Q_2 \\ P^{\prime}_4(-4,-5)\rightarrow P^{\doubleprime}_1(-(-4),-5)\rightarrow P^{\doubleprime}_1=(4,-5)=Q_4 \end{gathered}$

7 0

1 year ago