Graph the line.<br> y = -5x + 2

Y_Kistochka [10]

31.9
hope this helps

3 0

3 years ago

The weight of tigers follow a normal distribution with a mean of 220 kg and a SD of 30 kg. 1) If we randomly select a tiger, wha

mylen [45]

Answer:

0.89736

Step-by-step explanation:

We solve this question using z score formula

Z score = x - μ/σ

x = raw score

μ = population mean

σ = population standard deviation

Hence,

x = 258, μ = 220, σ = 30

Z = 258 - 220/30

=1.26667

Probability value from Z-Table:

P(x<258) = 0.89736

Therefore, the probability that his weights is less than 258 kg is 0.89736

6 0

3 years ago

The intersection of 2 planes can be defined as a_____?

tamaranim1 [39]

The intersection of two planes is a line. That's not necessarily the definition of a line, but that's probably what they're looking for.

Answer: line

5 0

2 years ago

Read 2 more answers

There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s

harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

Original Mean = $70,000

$\frac{\sum X}{n}$ = $70,000 ⇒ $\sum X$ = 70,000 * 10 = $700,000

New $\sum X$ after change = $700,000 - $100,000 + $1,000,000 = $1600000

Therefore, New mean = $\frac{1600000}{10}$ = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = $20000^{2}$ i.e. $20000^{2}$ = $\frac{\sum x^{2} - n*xbar }{n -1}$

Original $\sum x^{2}$ = ( $20000^{2}$ * (10-1)) + (10 * 70,000) = $3,600,700,000

New $\sum x^{2}$ = $3,600,700,000 - $100,000^{2} + 1,000,000^{2}$ = 9.936007 * $10^{11}$

New Variance = $\frac{new\sum x^{2} - n*new xbar }{n -1}$ = $\frac{9.936007 *10^{11} - 10*160000 }{10 -1}$ = 1.103999 * $10^{11}$ Therefore, standard deviation after change = $\sqrt{1.103999 * 10^{11} }$ = $332264.804 .