Can someone help me with this question

attashe74 [19]

Problem 1

Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.

Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.

With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.

Refer to the diagram below.

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Problem 2

I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.

In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.

8 0

2 years ago

The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

<em>The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.</em>

5 0

3 years ago

What is the image point of (-1,1) after a translation left 2 units and<br> down 2 units?

AveGali [126]

Answer:

The new point is (-3,-1)

Step-by-step explanation:

yessir

3 0

3 years ago

C(n) = 4/9(-3)^n-1<br> What is the 3rd term in the sequence?

Yanka [14]

Step-by-step explanation:

$c(n) = \frac{4}{9} ( - 3)^{n - 1} \\ \\ \therefore \: c(3) = \frac{4}{9} ( - 3)^{3- 1} \\ \\ \therefore \: c(3) = \frac{4}{9} ( - 3)^{2} \\ \\ \therefore \: c(3) = \frac{4}{9} \times 9 \\ \\ \huge{ \purple{ \boxed{\therefore \: c(3) = 4}}}$

8 0

3 years ago

A garden has width of <img src="https://tex.z-dn.net/?f=%5Csqrt13" id="TexFormula1" title="\sqrt13" alt="\sqrt13" align="absmidd

inessss [21]

Answer:

perimeter of the garden=2(length+width)

=2(7√13+√13)

=2(8√13)

=16√13

4 0

3 years ago

Can someone help me with this question ​

Can someone help me with this question