If a number is at most 5, which graph represents this number?

PLS HELP WILL GIVE BRAINIEST! You are flying in an airplane that has 2 engines. The engines individually have a success rate of

Salsk061 [2.6K]

Answer:

.01%*

Step-by-step explanation:

Individually, they both have a rate of failure of 1%.

.01 x .01 = .0001

Therefore they have a combined probability of failure of .01%

*Disclaimer!!

I am not the best at this area in math so I would still reference against someone else's answer, but this is my best effort :)

4 0

3 years ago

Suppose A and B are independent events. If P(A) = 0.6 and P(B) = 0.85, what is P(A and B)? A. .34 B. .51 C. .17 D. .68

s2008m [1.1K]

Answer:

0.51

Step-by-step explanation:

Mathematically, for two individual events;

P(AnB) = P(A) * P(B)

P(A n B) = 0.6 * 0.85 = 0.51

8 0

2 years ago

Read 2 more answers

Graph a line with a slope of.....

photoshop1234 [79]

Answer:

use google graph:)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

How long can late payments stay on your credit report?

ratelena [41]

The exception is if you are 60 days late often which will certainly hurt your scores. Otherwise, one late payment should not cause long term damage. 90 dayslate: This record will damage your credit scores significantly for up to seven years.

5 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)"

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.

7 0

3 years ago