All categories

3 years ago

The vertices of a triangle are located at L(4, –3), M(1, –3), and N(1, 0). What is the approximate perimeter of triangle LMN?

Mathematics

1 answer:

adelina 88 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

you need to find the distance between points L, M, N to know the length of sides LM, MN, NL

use the distance formula

d= √(x2-x1)² +(y2-y1)², where the points are (x1, y1) and (x2, y2)

LM = √(4-1)² + (-3- -3)² = √3² + 0² = 3

MN = √(1-1)² + (-3-0)² = √0² + 9 = 3

NL = √(4-1)² + (-3-0)² = √3² + 9 = √18 = 3√2

The perimeter is LM+MN+NL = 3+3+3√2 = 6+3√2 ≈ 10.243

You might be interested in

What is the value of X in degrees?

laiz [17]

3,50 degrees hope This heal

3 0

3 years ago

A Create a linear expression

frez [133]

Answer:

(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB. Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .Step-by-step explanation:

3 0

2 years ago

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

7 0

3 years ago

What is the total number of points scored by a player who makes 3 three-pointers, 5 two-pointers, and 6 free throws?

PolarNik [594]

25 because 3 threes is 9 and 5 twos is 10 so thats 19 plus 6 free throws which are 1 point so a total of 6

6 0

3 years ago

Read 2 more answers

Carlos is analyzing the results of an experiment on two groups of mice in which he compared the difference in their weight gain

Arlecino [84]

Given:difference in the mean weight gain is 0.60 gramsstandard deviation of the difference in sample mean is 0.305
68% confidence interval for the population mean difference is a) 0.305
0.60 + 1 * 0.3050.60 + 0.305 = 0.9050.60 - 0.305 = 0.295
95% confidence interval for the population mean difference is c) 0.61
0.60 + 2 * 0.3050.60 + 0.61 = 1.210.60 - 0.61 = -0.01

3 0

3 years ago