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1 year ago

What is the common ratio in this recursive definition ? HELP PLEASE

Mathematics

1 answer:

Sonbull [250]1 year ago

6 0

Answer:

(d) r = -2.1

Step-by-step explanation:

The common ratio is the coefficient of the a[n-1] term. It is the value that the previous term is multiplied by in order to get the next term.

r = -2.1

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(please zoom / zoom out if needed to see question. (: )

eduard

You can find the area of the rectangle and the triangle separately and then add them together which would result in 330in. I personally do not think you should use the trapezoid formula even though they said it is a trapezoid (a trapezoid is usually not shaped like that) but if you want to it would result in 270in.

3 0

2 years ago

The manufacturer of an electronic device claims that the probability of the device failing during the warranty period is 0.005.

Zielflug [23.3K]

Answer:

The value is $P(X \ge 15) = 0.5$

Step-by-step explanation:

From the question we are told that

The probability of the device failing during the warranty period is $p = 0.005$

The sample size is $n = 3000$

The random variable considered is x = 15

Generally this is distribution is binomial given the fact that there is only two out comes hence

X which is a variable representing a randomly selected selected electronic follows a binomial distribution i.e

$X \~ \ B(n , p)$

Now the mean is mathematically evaluated as

$\mu = n * p$

=> $\mu = 3000 * 0.005$

=> $\mu =15$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np(1 -p )}$

=> $\sigma = \sqrt{3000 * 0.005 * (1 - 0.005 )}$

=> $\sigma = 3.86$

Now given that n is very large, then it mean that we can successfully apply normal approximation on this binomial distribution

$P(X \ge 15) = P( \frac{X - \mu}{\sigma } \ge \frac{x - \mu}{\sigma } )$

Now applying Continuity Correction we have

$P(X \ge (15-0.5)) = P( \frac{X - \mu}{\sigma } > \frac{(15 -0.5) - 15}{3.86 } )$

Generally $\frac{X - \mu}{\sigma } = Z (The \ standardized \ value \ of \ X)$

$P(X \ge (15-0.5)) = P(Z >-0.130 )$

From the z-table

$P(Z >-0.130 ) =0.5$

Thus

$P(X \ge 15) = 0.5$

7 0

3 years ago

What are the lengths of line segments AB and BC

jasenka [17]

I need more information

4 0

3 years ago

Candice is preparing for her final exam in Statistics. She knows she needs an 65 out of 100 to earn an A overall in the course.

sertanlavr [38]

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

i.e., 68% values are between Mean Standard Deviation (SD).

Scores between 63 and 81 :

Note that 72 - 9 = 63 and

72 + 9 = 81

This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

Now, Scores between 63 and 81 =

= 68 X 12 = 816.

That means 816 students have scored between 63 and 81.

c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

i.e., 95% values are between Mean 2(SD).

Note that 90 = 72 + 2(9) = 72 + 18

Also, 54 = 63 - 18.

Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to of the values.

Therefore, Scores between 72 and 90 =

= 570.

That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

In this case, 5 % of the values lie between below 54 and above 90.

Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 =

= 2.5 X 12 = 30.

That is, 30 students have scored below 54.

Step-by-step explanation:

8 0

3 years ago

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Can someone solve this

navik [9.2K]

Answer:

482.84 cm

Step-by-step explanation:

8 0

2 years ago