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3 years ago

Tommy Thomas’s tankard holds 480ml when it is one quarter empty. How much does it hold when it is one quarter full?

Mathematics

1 answer:

agasfer [191]3 years ago

6 0

Answer:

Tommy Thomas's tankard holds 160ml when it is one-quarter full.

Step-by-step explanation:

When Tommy's tankard holds 480ml when it's one-quarter empty. There are three other quarters in the tankard. So, you would divide 480 by 3 to see how much is in each of the other 3 quarters. The answer comes down to 160ml per quarter, which equals one-quarter full.

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Solve the equation -3(x - 14) + 9x = 6x + 24. Does the equation have

Black_prince [1.1K]

-3(x-14)+9x=6x+24
-3x+42+9x=6x+24
-3x+9x-6x=-24-42
0=-18

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3 years ago

Do anybody play valorant?

BartSMP [9]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the value of x? Enter your answer in the box. x =

FinnZ [79.3K]

Answer:

x = 12

Step-by-step explanation:

IMPORTANT: Both answers should add up to 180 if they dont you did it wrong!

(10x - 20) + (6x + 8 ) = 180 plug in 12 for both of them

(10 * 12 - 20) + 6 * 12 + 8) = 120

(120 - 20) + (72+ 8 ) = 180

(100) + (80) = 180

so x = 12

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3 years ago

On a flight from New York to London, an airplane travels at a constant speed. An equation relating the distance traveled in mile

lions [1.4K]

Complete question :

On a flight New York to London an airplane travels at a constant speed. An equation relating the distance traveled in miles d to the number of hours flying t is t= 1/500d. How long will it take the airplane to travel 800 miles?

Answer:

1.6 hours

Step-by-step explanation:

Given the function :

Time t taken for flying distance d is given by:

t = 1/500 * d

Tine taken to fly 800 miles;

d = 800 miles

t = 1/500 * (800)

t = 800 / 500

t = 1.6

Hence, time t = 1.6 hours

3 0

3 years ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence 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:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

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

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

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

The statistic to check the hypothesis is given by:

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

Part b

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:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

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

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

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

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

"=1-CHISQ.DIST(33.75,2,TRUE)"

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

7 0

3 years ago