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PolarNik [594]
3 years ago
15

Two variables have a positive non-linear correlation. Does the independent variable increase or decrease as the dependent variab

le increases
Mathematics
1 answer:
sergey [27]3 years ago
3 0

Answer:

The answer is "Cannot determine from the information given".

Step-by-step explanation:

Please find the image file in the attachment.

Its positive correlation is nothing but the dependent share changes whenever we increase its different variables, and then we can say the situation is non-linear, though decrease or increase can occur. Likewise, the dependent variable as well decreases because once we decrease the independent variable.

You might be interested in
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?
Trava [24]

Answer:

\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}

Step-by-step explanation:

The explicit formula for a geometric sequence:

a_n=a_1r^{n-1}

a_n - n-th term

a_1 - first term

r - common ratio

r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}

We have

a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...

The common ratio:

r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}

<h2>It's not a geometric sequence.</h2>

If a_3=-\dfrac{1}{16} then the common ratio is r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}

Put to the explicit formula:

a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}

Put a_n=\dfrac{1}{1024} and solve for <em>n </em>:

-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1

\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6

5 0
3 years ago
Eight packages of strawberries cost $24. How many packages can you buy for $12
romanna [79]
4 packages, because 24/12=2, 8/2=4 packages.
8 0
3 years ago
For brainiest:):):):):):):):)
Oksi-84 [34.3K]

Answer:

(a) 0.6

(b) 0.4

Step-by-step explanation:

If I am reading the little table right, You have made 9 shots, and have missed 6 of them. Therefore, in total, you have made 9+6=15 shots.

The definition of probability is the number of events of a desired outcome divided by the total number of outcomes. Since you wish to make a shot for part a, the number of desired outcomes thus far is 9, out of 15 total shots. Thus,

P(making a shot)=\frac{9}{15}=\frac{3}{5}=0.6

Since making a shot and not making a shot are mutually exclusive events, and the only two events that could occur when shooting a basket, the probability of not making a shot and the probability of making a shot are all of the possible outcomes, and is thus equal to 1.

P(making a shot)+P(missing a shot) =1

P(missing a shot)=1-P(making a shot)

P(missing a shot)=1-0.6=0.4

QED

3 0
3 years ago
How to prove that square root of x where x is rational is rational?
Verdich [7]

In general, you cannot prove that because it is not true.

2 is rational. √2 is not.

5 0
3 years ago
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