Why cant i see the option to ask for a tutor:( is night time that's why? please answer

Marla started her school work at 8:48am. If she has to do 4 hours of school, what time will she be off?

grigory [225]

Answer:

See below.

Step-by-step explanation:

Q) 8:48 a.m. + 4 hours

= 12:48 p.m.

Bonus Q) 8:56 a.m. + 4 hours

= 12:56 p.m.

They will both be off by 12:56 p.m.

8 0

2 years ago

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the cost of a candle varies directly with the volume of the candle. If a candle with a volume of 80cm^3 costs $12, what would be

ivanzaharov [21]

First, you need to find out how much will the candle cost per volume. If a candle with a volume of 80cm^3 costs $12, the price of the candle per cm^3 would be: $12/80cm^3= $0.15/cm^3

Then the price of a candle with 120 cm^3 volume would be:120cm^3* $0.15/cm^3= $18

4 0

3 years ago

Suppose each license plate in a certain state has three letters followed by three digits. The letters and and the digits , , and

exis [7]

Answer:

17576000 different possible license plates

Step-by-step explanation:

Note that in the question, repetition is allowed for both the letters and digits.

For the first letter, there are 26 choices to make. For the each of the letters, there are 26 choices for the second letter, and for each of the letters, there are another 26 choices for the third letter.

Therefore, this gives us 26x26x26 = 17576 possible pairs of letters.

For the three numbers, we have ten possibilities for the first digit, ten possibilities for the second digit, and ten possibilities for the third digit.

Therefore, this gives us 10x10x10 =1000 possible numbers.

The combined possibility of the license plates in both three letters and three numbers is 17576 x 1000 = 17576000 different possible license plates.

7 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

4 years ago

What is the rule of this pattern 1, 21, 85, 341

weqwewe [10]

5 x 4 +1 =21

21 x 4 +1 = 85

85 x 4 + 1 = 341

4 0

3 years ago

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Why cant i see the option to ask for a tutor:( is night time that's why? please answer ​

Why cant i see the option to ask for a tutor:( is night time that's why? please answer