A regular octagon has what type of symmetry?

Find the first 5 terms of the arithmetic sequence : a(1)=-3 , a(17)=-20

madreJ [45]

The explicit formula for arithmetic sequence is:
an=a+(n-1)d
where:
a=first term
d=common difference
given:
a3=22
a(17)=-20
substituting this in our equation we get:
22=a+(3-1)d
22=a+2d
a=22-2d........i

also:
-20=a+(17-1)d
-20=a+16d.....ii
but substituting i in ii we get:
-20=22-2d+16d
-20-22=14d
-42=14d
d=-3
but:
a=22-2d
a=22-2(-3)
a=28
thus the formula will be:
an=28-3(n-1)
thus the first term will be 28
the 2nd term will be:
a2=28-3(2-1)
a2=25

the 3rd term will be:
a3=28-3(3-1)
a3=28-6
a3=22

a4=28-3(4-1)
a4=28-9
a4=15

a5=28-3(5-1)
a5=28-3(4)
a5=28-12
a5=15

5 0

2 years ago

The authors of a paper concluded that more boys than girls listen to music at high volumes. This conclusion was based on data fr

schepotkina [342]

Answer:

Since the calculated value of z= 2.82 does not lie in the critical region the null hypothesis is accepted and it is concluded that the sample data support the authors' conclusion that the proportion of the country's boys who listen to music at high volume is greater than this proportion for the country's girls.

The value of p is 0 .00233. The result is significant at p < 0.10.

Step-by-step explanation:

1) Let the null and alternate hypothesis be

H0: μboys − μgirls > 0

against the claim

Ha: μboys − μgirls ≤ 0

2) The significance level is set at 0.01

3) The critical region is z ≤ ± 1.28

4) The test statistic

Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]

Here p1= 397/768= 0.5169 and p2= 331/745=0.4429

pc = 397+331/768+745

pc= 0.4811

qc= 1-pc= 1-0.4811=0.5188

5) Calculations

Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]

z= 0.5169-0.4429/√ 0.4811*0.5188( 1/768+ 1/745)

z= 2.82

6) Conclusion

Since the calculated value of z= 2.82 does not lie in the critical region the null hypothesis is accepted and it is concluded that the sample data support the authors' conclusion that the proportion of the country's boys who listen to music at high volume is greater than this proportion for the country's girls.

7)

The value of p is 0 .00233. The result is significant at p < 0.10.

8 0

2 years ago

Y = –6x+2

qwelly [4]

The answer is Infinate Solutions I believe we are working on these types of equations in school rn

5 0

3 years ago

Read 2 more answers

The sum of three consecutive integers is 168. What are the integers

Tema [17]

Answer:

3X + 3 = 168

3X + 3 - 3 = 168 - 3

3X = 165

3X/3 = 165/3

X = 55

Step-by-step explanation:

We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 168. Therefore, you can write the equation as follows: (X) + (X + 1) + (X + 2) = 168

3 0

2 years ago

Suppose you play a game in which 5 dice (6-sided) are rolled simultaneously.

goblinko [34]

The expected value of the game in which 5 dice (6-sided) are rolled simultaneously is -0.2094.

<h3>How to find the mean (expectation) and variance of a random variable?</h3>

Supposing that the considered random variable is discrete, we get:

Mean = $E(X) =$ $\sum_{\forall x_i} f(x_i)x_i$

Here, $x_i; \: \: i = 1,2, ... ,$ n is its n data values and $f(x_i)$ is the probability of $X = x_i$

Suppose you play a game in which 5 dice (6-sided) are rolled simultaneously.

If a "4" is rolled, then you win $2 for each "4" showing.
If all the dice are showing "4", you win $1000.
If none of the dice are showing "4", then you lose $5.

Let Y is the amount of money player won. The value of X can be,

$Y=2,4,6,8,1000,-5$

<h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consists of n independent Bernoulli trials. Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

The expected value and variance of X are:

$E(X) = np\\ Var(X) = np(1-p)$

Put the values as 5 trials for each time 4 appears.

$P(X =0) = \: ^5C_1(\dfrac{1}{6})^0(1-\dfrac{1}{6})^{5-0}=0.4018 \:\\P(X =1) = \: ^5C_1(\dfrac{1}{6})^1(1-\dfrac{1}{6})^{5-1}=0.402\\P(X =2) = \: ^5C_2(\dfrac{1}{6})^2(1-\dfrac{1}{6})^{5-2}=0.162\\P(X =3) = \: ^5C_3(\dfrac{1}{6})^3(1-\dfrac{1}{6})^{5-3}=0.032\\P(X =4) = \: ^5C_4(\dfrac{1}{6})^4(1-\dfrac{1}{6})^{5-4}=0.0032\\P(X =5) = \: ^5C_5(\dfrac{1}{6})^5(1-\dfrac{1}{6})^{5-5}=0.00013\\$