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Elanso [62]
3 years ago
10

What is the 9th term of the geometric sequence 4, −20, 100, …?

Mathematics
2 answers:
aksik [14]3 years ago
8 0
1 = 4
2 = -20
3 = 100
4 = -500
5 = 2500
6 = - 12,500
7 = 62, 500
8 = - 312, 500
9 = 1,562,500

Hope this helped☺☺
blagie [28]3 years ago
4 0
4, -20, 100, -500, 2500, -12500, 62500, -312500, 1562500.
1,562,500 is your answer. 
I hope this helps!
(All the numbers were multiplied by -5 to get the answer.)
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If a vector's magnitude is 36 units right and 33 units up, what's the approximate magnitude of a vector that's one-third the len
omeli [17]

Answer:

The correct option is B. 16.3

Step-by-step explanation:

We are given a vector with components

V = 36x +33y

To solve this problem, we find its magnitude

|V| = SQRT (36^2 + 33^2)

|V| = SQRT (2385)

|V| = 48.836

To find the magnitude of the new vector we divide by three the previous quantity

|V2| = 48.836/3 = 16.278

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Mitch refuses to dine at a certain restaurant because he recalls reading two or three negative reviews of its service on a popul
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In this situation, we can actually see that Mitch was simply basing her taste on the information that is readily available to her. She does not even consider whether those reviews are legit or reliable. Mitch is falling into what we call as:

 

“the availability heuristic”

4 0
3 years ago
Help if possible.
skelet666 [1.2K]
A ) Length = 5 - 2 x
Width = 3 - 2 x
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Step-by-step explanation:

6 0
3 years ago
Using the simple random sample of weights of women from a data​ set, we obtain these sample​ statistics: nequals45 and x overbar
Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : n=45

\overline{x}=148.79\text{ lb}

\sigma=31.37\text{ lb}

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = \mu=148.79\text{ lb}

b) The confidence interval for the population mean is given by :-

\mu\ \pm E, where E is the margin of error.

Formula for Margin of error :-

z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given : Significance level : \alpha=1-0.90=0.1

Critical value : z_{\alpha/2}=z_{0.05}=\pm1.645

Margin of error : E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69

Now, the 90% confidence interval for the population mean will be :-

148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)

Hence, the 90​% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0
3 years ago
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