NEEDS HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The passing yards for the top 5 quarterbacks in the country are 3,832, 3,779, 3,655, 3,642, and 3,579. Find the variance and sta

Leviafan [203]

First get the average passing yards:

(3,832 + 3,779 + 3,655 + 3,642 + 3,579) / 5 = 3,697.4

Now get the squared residuals for each quarterback's passing yards. That is, compute the difference between each data and the average, then square the result. For example,

(3,832 - 3,697.4)^2 = 134.6^2 = 18,117.2

For the others, you should get squared residuals of

6,658.56

1,797.76

3,069.16

14,018.6

Take the sum of the squared residuals, then - since this is a sample, and not a population of all quarterbacks - divide the sum by 5 - 1 = 4 to get the variance:

(18,117.2 + 6,658.56 + 1,797.76 + 3,069.16 + 14,018.6)/4 = 10,915.3

The standard deviation is just the square root of the variance:

√(10,915.3) ≈ 104.48

5 0

3 years ago

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21-3+18÷6=6<br> Determine if True?

Lunna [17]

Answer:

False

Step-by-step explanation:

$21-3+18\div 6 \\\\=21-3+3\\\\=21-0\\\\=21\neq 6$

7 0

2 years ago

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Which description best fits the graph

Masteriza [31]

Decreasing then increasing, y goes down then y goes back up

7 0

3 years ago

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A regression model involved 18 independent variables and 200 observations. The critical value of t for testing the significance

Evgen [1.6K]

Answer:

The correct answer to the following question will be Option d (181 degrees of freedom).

Step-by-step explanation:

The given values are:

Regression model,

n = 200

Observations,

p = 18

Now,

⇒ $n-p-1$

On putting the estimated values, we get

⇒ $200-18-1$

⇒ $181$

So that the correct choice will be "181 degrees of freedom".

6 0

3 years ago

help solve this problem Based on a poll, 62% of Internet users are more careful about personal information when using a public

erma4kov [3.2K]

Answer

The probability that among four randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot is 0.979.

Step-by-step explanation:

Since 62% of Internet users are more careful about personal information when using a public Wi-Fi hotspot.

Therefore, the Probability of 62% of Internet users are more careful about personal information when using a public Wi-Fi hotspot is, p=0.62.

We have to find the probability that at least one user is more careful about personal information when using wi-fi hotspot.

Probability of internet users are not more careful about personal information when using a public wi-fi hotspot, i.e $q=1-p$

$q=1-0.62=0.38$

Use Binomial Distribution, i.e, $P\left ( X \right )=^nC_xp^xq^\left ( n-x \right )$

p=0.62, q=0.38, n=4.

Required Probability $P\left ( X\geq 1 \right )=P\left ( X=1 \right )+P\left ( X=2 \right )+P\left ( X=3 \right )+P\left ( X=4 \right )$

or

$P\left ( X\geq 1 \right )=1-P\left ( X=0 \right )$

=1- $^4C_0p^0q^\left ( 4-0 \right )$

=1- $4\cdot1\cdot q^\left ( 4-0 \right )$

=1- $4\cdot \left ( 0.38 \right )^4$

=1-0.02085136=0.97914864

Therefore, the probability that at least one users more careful about personal information when using wi-fi hotspot, i.e 0.979(approx).

RESULT:

<u>The result should not be impacted by this because volunteers are likely to have the most relevant responds.</u>

6 0

3 years ago