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ziro4ka [17]
3 years ago
10

Let’s work with some actual data this time. Go online to research and find a set of real-world data in two variables. The data c

an be in tabular form or in a scatter plot. Choose data that has a reasonable number of data points so you’re able to uncover trends. For the purposes of this activity, the data must not show a linear association. Describe the data you’ve identified in a sentence or two, and include a link to the data.
Mathematics
2 answers:
HACTEHA [7]3 years ago
6 0

Answer:

First u have to add the var

Step-by-step explanation:

4vir4ik [10]3 years ago
6 0

Answer:

well you gotta dp this yourself but ijust made up data between nike and jordans

Step-by-step explanation:

You might be interested in
Tom is playing football . He carries the ball 7 1/3 yards and then goes backwards 2 feet how far did Tom Cary the ball in feet a
Katen [24]

Answer:

20 feet

Step-by-step explanation:

convert yard to foot

1 yard = 3 foot

7 1/3 x 3 = 22

22 -2 = 20 feet

4 0
3 years ago
Transform both equations in each system of equations so that each coefficient is an integer.
inysia [295]
For the first one multiply each term by the LCM of 4 and 6, which is 12:-

15x -36 = 2y..................(1)

and second multiply by 5:-

2x + y = 9..............(2)

solving:-
from equation (2) y = 9-2x,  substitute  in equation (1):-

15x - 36 = 2(9-2x) = 18 - 4x
19x = 54
x = 54/19 = 2.84

so y =  9 - 2(2.84) = 3.32


4 0
3 years ago
Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are ap
miss Akunina [59]

Answer:

a) The probability of Jack scoring higher is 0.3446

b) They probability of them scoring above 350 is 0.2119

Step-by-step explanation:

Lets call X the random variable that determines Jill's bowling score and Y the random variable that determines jack's. We have

X \simeq N(170,400)\\Y \simeq N(160,225)

Note that we are considering the variance on the second entry, the square of the standard deviation.

If we have two independent Normal distributed random variables, then their sum is also normally distributed. If fact, we have this formulas:

N(\lambda_1, \sigma^2_1) + N(\lambda_2, \sigma^2_2) = N(\lambda_1 + \lambda_2,\sigma^2_1 + \sigma^2_2) \\r* N(\lambda_1, \sigma^2_1) = N(r\lambda_1,r^2\sigma^2_1)  

for independent distributions N(\lambda_1, \sigma^2_1) , N(\lambda_2, \sigma^2_2) , and a real number r.

a) We define Z to be Y-X. We want to know the probability of Z being greater than 0. We have

Z = Y-X = N(160,225) - N(170,400) = N(160,225) + (N(-170,(-1)^2 * 400) = N(-10,625)

So Z is a normal random variable with mean equal to -10 and vriance equal to 625. The standard deviation of Z is √625 = 25.

Lets work with the standarization of Z, which we will call W. W = (Z-\mu)/\sigma = (Z+10)/25. W has Normal distribution with mean 0 and standard deviation 1. We have

P(Z > 0) = P( (Z+10)/25 > (0+10)/25) = P(W > 0.4)

To calculate that, we will use the <em>known </em>values of the cummulative distribution function Φ of the standard normal distribution. For a real number k, P(W < k) = Φ(k). You can find those values in the Pdf I appended below.

Since Φ is a cummulative distribution function, we have P(W > 0.4) = 1- Φ(0.4)

That value of Φ(0.4) can be obtained by looking at the table, it is 0.6554. Therefore P(W > 0.4) = 1-0.6554 = 0.3446

As a result, The probability of Jack's score being higher is 0.3446. As you may expect, since Jack is expected to score less that Jill, the probability of him scoring higher is lesser than 0.5.

b) Now we define Z to be X+Y Since X and Y are independent Normal variables with mean 160 and 170 respectively, then Z has mean 330. And the variance of Z is equal to the sum of the variances of X and Y, that is, 625. Hence Z is Normally distributed with mean 330 and standard deviation rqual to 25 (the square root of 625).

We want to know the probability of Z being greater that 350, for that we standarized Z. We call W the standarization. W is s standard normal distributed random variable, and it is obtained from Z by removing its mean 330 and dividing by its standard deviation 25.

P(Z > 350) = P((Z  - 330)/25 > (350-330)/25) = P(W > 0.8) = 1-Φ(0.8)

The last equality comes from the fact that Φ is a cummulative distribution function. The value of Φ(0.8) by looking at the table is 0.7881, therefore P(X+Y > 350) = 1 - Φ(0.8) = 0.2119.

As you may expect, this probability is pretty low because the mean value of the sum of their combined scores is quite below 350.

I hope this works for you!

Download pdf
6 0
3 years ago
Help! Pls pls pls! Fast!
pickupchik [31]

it is transformed |x| function. moved down by and right by 1 unit,

so $y=|x-1|-1$

4 0
3 years ago
Please help I need help on this
natulia [17]

Answer: the answer is 120 for it the area but that is not on the answer list.

Step-by-step explanation:

7 0
3 years ago
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