Question 4 pls thanks, I will mark Brainliest

yaroslaw [1]

Answer:A)0.5 We can see in the graph , that it is bell-shaped along x =2. A bell-shaped graph along one value is called symmetric graph and it represents a normal distribution.

Since, the give graph is symmetric around x=2, so the mean of the data is 2.

The point immediate left to the mean represents x-σ

so,

2 - σ = 1.5

So,

σ = 0.5

The sigma represents standard deviation.

Hence, Option A is correct ..

Step-by-step explanation:and what is on d is 2.5

6 0

3 years ago

We all hate to bring change to a store. By using random numbers, we can eliminate the need for change and give the store and the

Sunny_sXe [5.5K]

Answer:

Round off to closest value.

Step-by-step explanation:

In order to remove the requirement for change, every time the client buys a product for $0.20 the random number that lies between 0 and 1 would be produced also the same would be rounded to the closest value i.e. 0 or 1. In the case when 0 arise so the customer would not have to pay any amount but if 1 arise so the customer have to pay $1. So this type of method would remove the requirement for a minor change

6 0

2 years ago

A graph is shown below

andrezito [222]

We have that

using a graph tool
see the attached figure

case A) x − 2y > 3
this inequality represented the graph

case B) x − 2y < 3
this inequality not represented the graph

case C) 2x − y > 3
this inequality not represented the graph

case D) 2x − y < 3
this inequality not represented the graph

the answer is
the option A, x − 2y > 3

4 0

3 years ago

Read 2 more answers

The quotient of two negative integers results in an integer. How does the value of the quotient compared the value of the origin

Taya2010 [7]

The first two were negative integers and the quotient is therefor a positive integer (Two negatives equal a positive)

5 0

3 years ago

Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that

Bond [772]

Taylor series is $f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }$

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

$f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }$

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

$f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }$

.

Following the pattern, we can see that for $f^{n}(x)$ ,

$f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} } \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}$

This applies for n ≥ 1, Expressing f(x) in summation, we have

$\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}$

Combining ln2 with the rest of series, we have

$f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }$

Taylor series is $f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }$

Find out more information about taylor series here

brainly.com/question/13057266

#SPJ4

3 0

2 years ago