-3x - 8 = 50<br> Anyone know how to solve this by elimination?

Assume that a procedure yields a binomial distribution with n=3 trials and a probability of success of p=0.300. use a binomial p

NNADVOKAT [17]

$\mathbb P(X=x)=\begin{cases}\dbinom3xp^x(1-p)^{3-x}&\text{for }0\le x\le3\\\\0&\text{otherwise}\end{cases}$

$\implies\mathbb P(X=2)=\dbinom32(0.3)^2(0.7)^1=0.189=18.9\%$

7 0

3 years ago

Pls help meh<br> i will give brainlyest

MissTica

Answer:

The answer is 75

can me be brainliest?

Step-by-step explanation:

PEMDAS

20x4-100/20

80-5

75

5 0

3 years ago

Read 2 more answers

The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament.

NeX [460]

Answer:

Step-by-step explanation:

Hello!

(Data and full text attached)

The stem and leaf plot is a way to present quantitative data. Considering two-digit numbers, for example 50, the tens digits are arranged in the stem and the units determine the leafs.

So for the stem and leaf showing the digs of the top players of the tournament, the observed data is:

41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97

n= 15

Note that in the stem it shows the number 8, but with no leaf in that row, that means that there were no "eighties" observed.

a) 6 Players had more than 60 digs.

b)

To calculate the mean you have to use the following formula:

X[bar]= ∑x/n= (41 + 41 + 43 + 43 + 45 + 50 + 42 + 53 + 54 + 62 + 63 + 63 + 67 + 75 + 97)/15= 849/15= 56.6 digs

To calculate the median you have to calculate its position and then identify its value out of the observed data arranged from least to greatest:

PosMe= (n+1)/2= (15+1)/2= 8 ⇒ The median is in the eight place:

41, 41, 43, 43, 45, 50, 42, 53, 54, 62, 63, 63, 67, 75, 97

The median is Me= 53

53 is the value that separates the data in exact halves.

The mode is the most observed value (with more absolute frequency).

Consider the values that were recorded more than once

41, 41

43, 43

63, 63

41, 43 and 63 are the values with most absolute frequency, which means that this distribution is multimodal and has three modes:

Md₁: 41

Md₂: 43

Md₃: 63

The Range is the difference between the maximum value and the minimum value of the data set:

R= max- min= 97 - 41= 56

c)

The distribution is asymmetrical, right skewed and tri-modal.

Md₁: 41 < Md₂: 43 < Me= 53 < X[bar]= 56.6 < Md₃: 63

Outlier: 97

d)

An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.

Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:

X < Q₁ - 1.5 IQR

X > Q₃ + 1.5 IQR

PosQ₁= 16/4= 4

Q₁= 43

PosQ₃= 16*3/4= 12

Q₃= 63

IQR= 63 - 43= 20

Q₁ - 1.5 IQR = 43 - 1.5*20= 13 ⇒ There are no values 13 and below, there are no lower outliers.

Q₃ + 1.5 IQR = 63 + 1.5*20= 93 ⇒ There is one value registered above the calculated limit, the last observation 97 is the only outlier of the sample.

The mean is highly affected by outliers, its value is always modified by the magnitude of the outliers and "moves" its position towards the direction of them.

Calculated mean with the outlier: X[bar]= 849/15= 56.6 digs

Calculated mean without the outlier: X[bar]= 752/14= 53.71 digs

I hope this helps.