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2 years ago

Help me to answer this question pls

Mathematics

1 answer:

attashe74 [19]2 years ago

8 0

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.

=====================================================

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.

You might be interested in

What is the solution to the system of equation below Y equals -1/3 X +6 and X equals -6

Evgen [1.6K]

Answer:

y = 8

Step-by-step explanation:

y= $\frac{1}{3} x$ + 6

$x$ = -6

y=( $\frac{1}{3}$ x 6) + 6

y= 2+6 = 8

6 0

3 years ago

Tanisha's website had x visitors on Friday. On Saturday, it had 500 more visitors than it did on Friday. The total

liq [111]

If the number of visitors on Friday, Saturday, and Sunday is x, x + 500, and x + 250. Then the total number of visitors will be 3x + 750.

<h3>What is Algebra?</h3>

Algebra is the study of mathematical symbols, and the rule is the manipulation of those symbols.

Tanisha's website had x visitors on Friday.

On Saturday, it had 500 more visitors than it did on Friday. That is x + 500

The total number of visitors on Friday and Saturday was twice the number on Sunday. That is (2x + 500)/2

Then the total number of the visitors will be

→ Friday + Saturday + Sunday

Then we have

$\rm \rightarrow x + x + 500 + \dfrac{2x + 500}{2}\\\\\rightarrow 2x + 500 + x + 250\\\\\rightarrow 3x + 750$

More about the Algebra link is given below.

brainly.com/question/953809

#SPJ1

6 0

2 years ago

Six stand-up comics, A, B, C, D, E, and F, are to perform on a single evening at a comedy club. The order of performance is

kiruha [24]

Answer:

a) 1/6

b) 1/36

c) 1/720

d) 1/3

Step-by-step explanation:

a) Any of the six comics can perform in the fourth place, so there is one chance in six that Comic F is the one that performs fourth.

$P(F=4th)=1/6$

b) In this case, we have two conditions. Both are independent of each other, so the probability of both happening is the product of the probabilities of each happening individually:

$P(D=4th \&B=2nd)=P(D=4th)\cdot P(B=2nd)=(1/6)\cdot(1/6)=1/36$

c) This combination is one in all possible orders of perform. The amount of combinations of orders is n!=6!=720 possible combinations. So the probability of this specific order is:

$P(F, D, B, A, C, E)=1/720$

d) In this case, of the six possible comics performing in the last place, we calculate the probability of 2 of them being in that place. So the probability is:

$P(C=6th\,or\, E=6th)=2/6=1/3$

8 0

3 years ago

12. Using the value of 0.538 for the proportion of those surveyed who said they exercise at least 30 minutes three or more days

ICE Princess25 [194]

Answer:

$ME=1.96\sqrt{\frac{0.538 (1-0.538)}{15235}}=0.00792$

The margin of error on this case is approximated $\pm 0.792\%$ who is very clos to the +1% reported by the Gallup.

Step-by-step explanation:

Assuming this previous info : "Results are based on telephone interview conducted as part of the Gallup-Healthways Well-Being Index survey June 1–30,2013, with a random sample of 15,235 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia.

For results based on the total sample of national adults, one can say with 95% confidence that the maximum margin of sampling error is ±1 percentage point."

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The estimated proportion on this case is $\hat p =0.538$ and n =15235, the margin of error is given by:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And if we replace we got:

$ME=1.96\sqrt{\frac{0.538 (1-0.538)}{15235}}=0.00792$

The margin of error on this case is approximated $\pm 0.792\%$ who is very clos to the +1% reported by the Gallup.

5 0

3 years ago

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

Step-by-step explanation:

Let's denote the events as follows:

<em>F</em> = The word free occurs in an email

<em>S</em> = The email is spam

<em>V</em> = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

3 years ago

Help me to answer this question pl​s

Help me to answer this question pls