Helppp !!! Mathematics

What are the different methods in finding the LCM

Nezavi [6.7K]

Answer:

4 methods are there

Step-by-step explanation:

Find the prime factorization of each number.
Write each number as a product of primes, matching primes vertically when possible.
Bring down the primes in each column.
Multiply the factors to get the LCM.

please mark me as brainlist

8 0

2 years ago

The _______ is the distance around a circle.

AnnyKZ [126]

The circumference is the correct.

It can be found using the formula,

$circumference = 2 \pi \: r$

3 0

2 years ago

Read 2 more answers

A chemistry teacher needs to mix a 20% salt solution with an 80% salt solution to make 20 qt of a 40% salt solution. How many qu

Svetlanka [38]

Answer:

6 2/3 qt 80%
13 1/3 qt 20%

Step-by-step explanation:

It is often convenient to solve a mixture problem by letting a variable represent the quantity of the higher-concentration contributor to the mix.

__

We can let x represent the number of quarts of 80% solution needed. Then (20-x) is the number of quarts of 20% solution needed. The amount of salt in the final mix is ...

0.80x +0.20(20-x) = 0.40(20)

0.60x = 0.20(20) . . . . . . . . subtract 0.20(20) and simplify

x = 20/3 = 6 2/3 . . . . . . . . . divide by 0.60; quarts of 80% solution

(20 -x) = 13 1/3 . . . . . . . . . . amount of 20% solution needed

The teacher should mix 6 2/3 quarts of 80% solution with 13 1/3 quarts of 20% solution.

3 0

2 years ago

This is a really hard question

Anastasy [175]

Answer:

D. 4x - 6

Step-by-step explanation:

f(x) = 4x - 2
f(x - 1) = 4(x - 1) - 2 = 4x - 4 - 2 = 4x - 6

Correct option is D

5 0

3 years ago

Read 2 more answers

Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin

Artemon [7]

"Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find $\bar{d}$ , $s_{d}$ , the t-test statistic, and the critical values to test the claim that $\mu_{d} = 0$ "

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find $\bar{d}$
You are asked to find the mean difference between the two variables, which is given by the formula:
$\bar{d} = \frac{\sum (x - y)}{n}$

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example:
 $\bar{d} = \frac{6}{8} = 0.75$ 

B) Find $s_{d}$
You are asked to find the standard deviation, which is given by the formula:
 $s_{d} = \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }$

These are the steps to follow:
1) Subtract the mean difference from each pair's difference
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
$s_{d} = \sqrt{ \frac{101.5}{8-1} }$
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
$t = \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}$

where SE = standard error is given by the formula:
$SE = \frac{ s_{d} }{ \sqrt{n} }$

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract $\mu_{d}$
6) divide by the standard error

In our example:
SE = 3.81 / √8
= 1.346

The problem gives us $\mu_{d} = 0$ , therefore:
t = [(9.75 - 9) - 0] / 1.346
= 0.56

D) Find $t_{\alpha / 2}$
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find $t_{\alpha / 2} = 1.895$ 

Since our t-value is less than $t_{\alpha / 2}$ we can reject our null hypothesis!!

7 0

3 years ago