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2 years ago

Find the pattern of 1,2,6,18 and state if its arithmetic or geometric

Mathematics

1 answer:

Gnesinka [82]2 years ago

6 0

Answer:

its a geometric sequence where each term is 3 times the previous term.

Step-by-step explanation:

{2,6,18,54,162,486,1458...}

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Can someone plz help me ASAP

Andrews [41]

Day 3 is double than day two judging by hours baked... it's 96 cookies

5 0

3 years ago

The total weight of samsiah and Julia is twice the weight of Mary. if samsiah weighs 54 kg and Mary weighs 60 kg, find the actua

Wewaii [24]

s + j = 2m

s = 54

m = 60

54 + j = 120

j = 120 - 54

j = 66

4 0

2 years ago

Find the next 3 terms in the geometric sequence -36, 6, -1, 1/6, ...

timama [110]

we are given

geometric sequence -36, 6, -1, 1/6, ...

first term is -36

$a_1=-36$

now, we can find common ratio

$r=\frac{6}{-36}$

$r=-\frac{1}{6}$

now, we can find nth term

$a_n=a_1(r)^{n-1}$

now, we can plug values

and we get

$a_n=36(-\frac{1}{6})^{n-1}$

now, we can find 5th term , 6th term, 7th term

fifth term:

$a_5=36(-\frac{1}{6})^{4}$

$a_5=\frac{1}{36}$

sixth term:

$a_6=36(-\frac{1}{6})^{5}$

$a_6=-\frac{1}{216}$

seventh term:

$a_7=36(-\frac{1}{6})^{6}$

$a_7=\frac{1}{1296}$

so, next terms are

$a_5=\frac{1}{36}$ , $a_6=-\frac{1}{216}$

, $a_7=\frac{1}{1296}$ .............Answer

6 0

3 years ago

Crash Hospital Costs. A study was conducts to estimate hospital costs for accident victims who wore seat belts. Twenty randomly

kvv77 [185]

Answer:

a) The 99% confidence interval would be given by (5004.168;12603.832)

b) For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I

Step-by-step explanation:

1) 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".

$\bar X =9004$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=5629 represent the sample standard deviation

n=20 represent the sample size

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=20-1=19$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,19)".And we see that $t_{\alpha/2}=2.86$

Now we have everything in order to replace into formula (1):

$9004-2.86\frac{5629}{\sqrt{20}}=5004.168$

$9004+2.86\frac{5629}{\sqrt{20}}=12603.832$

So on this case the 99% confidence interval would be given by (5004.168;12603.832)

3) Part b

For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I

7 0

3 years ago

The midpoint of \overline{\text{AB}}AB is M(-6, -4)M(−6,−4). If the coordinates of AA are (-4, -3)(−4,−3), what are the coordina

valentina_108 [34]

Answer:

The coordinates of B is (-8,-5).

Step-by-step explanation:

The midpoint of line AB is M. The coordinate of M is (-6,-4).

The coordinates of A is (-4,-3)

We need to find the mid point of B.

If M(x,y) is the midpoint of the coordinates (x₁,y₁) and (x₂,y₂). The mid point theorem is used as follows :

$M(x,y)=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

Let the mid point of B is (x₂,y₂). Put (x,y) = (-6,-4), (x₁,y₁) = (-4,-3).

$(-6,-4)=(\dfrac{-4+x_2}{2},\dfrac{-3+y_2}{2})\\\\\dfrac{-4+x_2}{2}=-6\ \text{and}\ \dfrac{-3+y_2}{2}=-4\\\\-4+x_2=-12\ \text{and}\ -3+y_2=-8\\\\x_2=-12+4\ \text{and}\ y_2=-8+3\\\\x_2=-8\ \text{and}\ y_2=-5$

So, the coordinates of B is (-8,-5).

6 0

3 years ago