2. Determine the number of all possible diagonals drawn for polygon having a. 7 sides b. 10 sides c. n - sides​

Answer:

the answer of r is 8 i hope it will help

Answer: 9 feet

Step-by-step explanation:

From the information given, we have already been given the equation which is x(x-3)=54. Therefore we will find the value of x which will be:

x(x-3)=54

x² - 3x - 54

x² - 9x + 6x - 54

x(x - 9) + 6(x - 9)

Therefore,

(x - 9) = 0

x = 0 + 9

x = 9

The length is 9 feet

The width will be:

x - 3 = 9 - 3 = 6 feet

Answer:

37.94

Step-by-step explanation:

the 9th percentile is equal to a zscore of -1.34

-1.34=(x-50)/9

x=37.94

Answer:

120Pi units squared

Step-by-step explanation:

π*12²*(360-60)/360

= π*144*300/360

= π*144*5/6

= π*720/6

= π*120

= 120π or 120Pi units squared

Answer:

120Pi units squared

Step-by-step explanation:

on edge

have attached the solution, couldn't solve 13 and 17 tho, if it helps, then do mark me the brainliest...

Btw, which class r u in?

Step-by-step explanation:

Answer:

15=2p+3

Step-by-step explanation:

Answer:

A a vertex

Step-by-step explanation:

When rays meet they form a point is formed known as the vertex.

Answer:

71 feet

Step-by-step explanation:

94×2=188

330-188=142

142÷2=71

Answer:

The answer is "840".

Step-by-step explanation:

Following are the number of ways in which selecting a team A  by 9 children:

[tex]= ^{9_{C_{3}}\\\\\\[/tex]

[tex]=\frac{9!}{3! \times 6!} \\\\=\frac{9\times 8\times 7\times 6!}{3 \times 2\times 1\times 6!}\\\\=\frac{9\times 8\times 7}{3 \times 2\times 1}\\\\=\frac{3\times 4\times 7}{1}\\\\=\frac{84}{1}\\\\=84[/tex]

Following are the number of ways in which selecting a team B  by remaining 6 children:

 [tex]= ^{6}_{C_{3}}[/tex]

[tex]= \frac{6!}{(3! \times 3!)}\\\\= \frac{6!}{(3\times 2\times 1 \times 3!)}\\\\= \frac{6\times 5 \times 4 \times 3!}{(3\times 2\times 1 \times 3!)}\\\\= \frac{ 5 \times 4 \times 3!}{3!}\\\\= 5 \times 4 \\\\=20[/tex]

Following are the number of ways in which selecting a team C  by remaining 3 children:

[tex]= ^{3}_{C_{3}}\\\\=\frac{3!}{3!}\\\\= 1[/tex]

Following are the number of ways in which making 3 teams by 9 children:

[tex]= \frac{(84 \times 20 \times 1)}{3!}\\\\= \frac{(84 \times 20 )}{6}\\\\= 14 \times 20\\\\= 280\\\\[/tex]

(Note: we've split by 3! Because it also is necessary to implement three teams between themselves)

Now 3 leagues have to be played. One is going to be run by each team.

That is the way it is

Different possible divisions

[tex]= 280 \times 3!\\\\= 280 \times (3 \times 2 \times 1)\\\\= 840[/tex]  

Answer:

q=4

p=2

Step-by-step explanation:

3p+2q=14

10p+6q=44

10(3p+2q=14)

3(10p+6q=44)

30p+20q=140-

30p+18q=132

2q=8

2q/2=8/2

q=4

3p+2*4=14

3p+8=14

3p=14-8

3p/3=6/3

p=2

hope this helps

Answer:

[tex]p=2\\q=4[/tex]

Step-by-step explanation:

One is given the following system of equations,

[tex]3p + 4q = 22\\\\10p + 12q = 68[/tex]

The fastest method to solve a system of equations is the method of elimination. This process is manipulating one of the equations, by multiplying or diving it by a value, such that one of the coefficients variables in the equation is the additive inverse of the like term in the other equation. That way, when one adds the equations, one of the variables cancels out. Then one can solve for the other term. Finally, one can back sovle by substituting the value of the solved variable into one of the equations and simplifying to find the value of the other variable.

[tex]3p + 4q = 22\\\\10p + 12q = 68[/tex]

Manipulate the first equation so that the variable (q) cancels

[tex](3p + 4q = 22) *(-3)\\\\10p + 12q = 68[/tex]

[tex]-9p + -12q = -66\\\\10p + 12q = 68[/tex]

Add the equations,

[tex]-9p + -12q = -66\\\\10p + 12q = 68[/tex]

[tex](10p-9p)+(-12q+12q)=(-66+68)[/tex]

Simplify,

[tex](10p-9p)+(-12q+12q)=(-66+68)[/tex]

[tex]p=2[/tex]

Backsovle for the variable (p). Substitute the values of (p) into one of the original equations. Then simplify and use inverse operations to solve for the variable (q).

[tex]3p+4q=22[/tex]

Substitute,

[tex]3(2)+4q=22[/tex]

Simplify,

[tex]3(2)+4q=22[/tex]

[tex]6+4q=22[/tex]

Inverse operations,

[tex]6+4q=22[/tex]

[tex]4q=16\\q=4[/tex]

Answer:

Step-by-step explanation:

Part A.....let P be the number of people that will show up.....so....

The total amount of broccoli needed (in ounces)  =    8P  ounces

Part B

32  =  8P       divide both sides by 8

4 = P            so.....4 people can be fed.....!!!

Step-by-step explanation:

Answer:

Angle formed by the sector measuring x% will be 126°.

Step-by-step explanation:

Since, sum of all sectors formed in a circle is 100%.

By adding the measures of all the sectors,

x + x + 21 + 9 = 100

2x + 30 = 100

2x = 70

x = 35%

Now we know sum of all the central angles formed at the center of a circle = 360°

Therefore, angle formed by x% = 360° × 35%

                                                    = [tex]\frac{360\times 35}{100}[/tex]

                                                    = 126°

Answer:

B.

Step-by-step explanation:

there is not much to explain.

a strong correlation means there is a direct connection.

so, a change in a causes immediately a change in b.

and positive means that the changes go in the same sign direction. increase => increase. decrease => decrease.

Answer:

probability[Number of 6 random sample do not have cavities] = 0.8

Step-by-step explanation

Given:

Number of student do not have cavities = 60%

Number of random sample = 9 children

Find:

Probability[Number of 6 random sample do not have cavities]

Computation:

n = 9

p = 60% = 0.6

P(At least 6)  

Probability[Number of 6 random sample do not have cavities] = 1 - P(Less than 6)  

Probability[Number of 6 random sample do not have cavities] = 1 - P(Less than or equal to 6)

Probability[Number of 6 random sample do not have cavities] = 0.8

Answer:

1326

Step-by-step explanation:

[tex]{52\choose2}=\frac{52!}{(52-2)!2!}=\frac{52!}{50!*2!}=1326[/tex]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that [tex]\mu = 273, \sigma = 100[/tex]

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that [tex]n = 30, s = \frac{100}{\sqrt{30}}[/tex]

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}[/tex]

[tex]Z = 0.88[/tex]

[tex]Z = 0.88[/tex] has a p-value of 0.8106

X = 257

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}[/tex]

[tex]Z = -0.88[/tex]

[tex]Z = -0.88[/tex] has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that [tex]n = 50, s = \frac{100}{\sqrt{50}}[/tex]

X = 289

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}[/tex]

[tex]Z = 1.13[/tex]

[tex]Z = 1.13[/tex] has a p-value of 0.8708

X = 257

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}[/tex]

[tex]Z = -1.13[/tex]

[tex]Z = -1.13[/tex] has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that [tex]n = 100, s = \frac{100}{\sqrt{100}}[/tex]

X = 289

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}[/tex]

[tex]Z = 1.6[/tex]

[tex]Z = 1.6[/tex] has a p-value of 0.9452

X = 257

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}[/tex]

[tex]Z = -1.6[/tex]

[tex]Z = -1.6[/tex] has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

C is actually 0.8904

for anybody else stuck on this wondering why cengage is telling you c is wrong

Answer:

A and F

Step-by-step explanation:

A and F both represent instances of division of 14/5

B represent multiplication

C represent the reciprocal of the problem, 5/14

D represent addition

Answer:

[tex](a)\ P(White) = \frac{1}{2}[/tex]

(b) 10 additional white balls

(c) 10 additional black balls

Step-by-step explanation:

Given

[tex]White = 5[/tex]

[tex]Black =3[/tex]

[tex]Red = 2[/tex]

Solving (a): P(White)

This is calculated as:

[tex]P(White) = \frac{White}{Total}[/tex]

[tex]P(White) = \frac{5}{5+3+2}[/tex]

[tex]P(White) = \frac{5}{10}[/tex]

[tex]P(White) = \frac{1}{2}[/tex]

Solving (b): Additional white balls, if [tex]P(White) = \frac{3}{4}[/tex]

Let the additional white balls be x

So:

[tex]P(White) = \frac{White+x}{Total+x}[/tex]

This gives:

[tex]\frac{3}{4} = \frac{5+x}{10+x}[/tex]

Cross multiply

[tex]30+3x = 20 + 4x[/tex]

Collect like terms

[tex]4x - 3x = 30 - 20[/tex]

[tex]x = 10[/tex]

Hence, 10 additional white balls must be added

Solving (c): Additional black balls, if [tex]P(White) = \frac{1}{4}[/tex]

Let the additional black balls be x

So:

[tex]P(White) = \frac{White}{Total+x}[/tex]

So, we have:

[tex]\frac{1}{4} = \frac{5}{10+x}[/tex]

Cross multiply

[tex]10+x = 5 * 4[/tex]

[tex]10+x = 20[/tex]

Collect like terms

[tex]x = 20 -10[/tex]

[tex]x = 10[/tex]

Hence, 10 additional black balls must be added

Answer:

3+ 0 = 3

( 4+7) + 3.0 × 1 = 4 + (7+3)

9 + 8 = 8 + 9

Answer:

1 71years

Step-by-step explanation:

x+10=81

x=71

2. x+8=49

x=41

3. x+11=100

x=99

Answer:

19.3 miles per gallon

Step-by-step explanation:

First, subtract 54,042.8-53,737.7. The answer is 305.1

Then, find the unit rate. 305.1/15.8

You get 19.31012658. The prompt says to round to the nearest tenth, so round, and you get 19.3.

That's your answer!

Answer:

The sales tax on a 320 dollars purchase is of $29.6.

Step-by-step explanation:

State sales tax y is directly proportional to retail price x.

This means that:

[tex]y = cx[/tex]

In which c is the constant of proportionality.

An item that sells for 156 dollars has a sales tax of 14.42 dollars.

This means that [tex]x = 156, y = 14.42[/tex]. We use this to find c. So

[tex]y = cx[/tex]

[tex]14.42 = 156c[/tex]

[tex]c = \frac{14.42}{156}[/tex]

[tex]c = 0.0924[/tex]

Then

[tex]y = 0.0924x[/tex]

What is the sales tax on a 320 dollars purchase?

y when [tex]x = 320[/tex]. So

[tex]y = 0.0924(320) = 29.6[/tex]

The sales tax on a 320 dollars purchase is of $29.6.

Answer:

the operating characteristics have been solved below

Step-by-step explanation:

we have an average of 10 minutes per customers

μ = mean service rate = 60/10 = 6 customers in one hr

the average number of customers that are waiting in line

mean arrival λ = 2.5

μ = 6

[tex]Lq = \frac{2.5^{2} }{6(6-2.5)} \\[/tex]

= 6.25/21

= 0.2976

we calculate the average number of customers that are in the system

[tex]L=Lq+\frac{2.5}{6}[/tex]

= 0.2976+0.4167

= 0.7143

we find the average time that a customer spends in waiting

[tex]Wq=\frac{0.2976}{2.5}[/tex]

= 0.1190 hours

when converted to minutes = 0.1190*60 = 7.1424 minutes

[tex]0.1190+\frac{1}{6}[/tex]

=0.2857

probability that arriving customers would wait for the service

= 2.5÷6 = 0.4167

Answer:

0.18431525 = 18.4%

Step-by-step explanation:

General Formula :

total trials Cn⋅p(success)^n⋅p(fail)^total−n

1198144* (.0218)^3 * (1-.0218)^191

Answer:

221

Step-by-step explanation:

Given sequence is ,

> 5 , 6 , 7 , 8 , 9.

The common difference is 6-5 = 1 .

Therefore , the 221st number will be

> 221 st term = 221 × 1 = 221 .

Hence the 221 st term is 221 .

Answer:

225

Step-by-step explanation:

d = 6 - 5 = 1 (common differences)

a = 5 (first term)

221st term

a+(n-1)d

5 +(221 - 1) 1

5 + 220 =225

Therefore the answer is 225

Answer:

a) 0.0147 = 1.47% probability that none of them graduates from the local community college.

b) 0.9398 = 93.98% probability that at most four will graduate from the local community college.

c) The expected number that will graduate is 2.85.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they will graduate, or they will not. The probability of a student graduating is independent of any other student graduating, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

57% of students who enter the college as freshmen go on to graduate.

This means that [tex]p = 0.57[/tex]

Five freshmen are randomly selected.

This means that [tex]n = 5[/tex]

a. What is the probability that none of them graduates from the local community college?

This is P(X = 0). So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{5,0}.(0.57)^{0}.(0.43)^{5} = 0.0147[/tex]

0.0147 = 1.47% probability that none of them graduates from the local community college.

b. What is the probability that at most four will graduate from the local community college?

This is:

[tex]P(X \leq 4) = 1 - P(X = 5)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{5,5}.(0.57)^{5}.(0.43)^{0} = 0.0602[/tex]

So

[tex]P(X \leq 4) = 1 - P(X = 5) = 1 - 0.0602 = 0.9398[/tex]

0.9398 = 93.98% probability that at most four will graduate from the local community college.

c. What is the expected number that will graduate?

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

In this question:

[tex]E(X) = 5*0.57 = 2.85[/tex]

The expected number that will graduate is 2.85.

it is 2000 ml after she drank it on Tuesday

Answer:

2000 mL

Step-by-step explanation:

1. The amount of milk left after 1/5 is drank is 4 litres, which is 4000 mL.

2. She drinks half of that, so divide 4000 mL by two.

2000 mL are left after tuesday