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Marizza181 [45]
2 years ago
5

Of Canada's total area of 9,976,140 km² , 755,170 km² is water. To the nearest tenth of percent, what part of Canada is water?

Mathematics
1 answer:
algol132 years ago
5 0

Answer:

755170/9976140 x 100

= 7.56976 %

= 8 % or (to the nearest 10th 10%) of Canada

Step-by-step explanation:

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Subtract. Fill in the missing numbers, 14 - 8 = 8= _____ + ________ So, 14 - 8 = _______
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Answer:

the answer is 6.

Step-by-step explanation:

I think they want you to skip count down by 4 and there's your anwser.

Hope I helped.

SJ

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3 years ago
The dance committee at Jefferson High School decides to charge students different prices for dance tickets depending on whether
valentina_108 [34]

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30

Step-by-step explanation:

4 0
3 years ago
Measure the length of the top of your desk in centimeters. Describe how you found the length.
Greeley [361]
Take a ruler and put it on your desk and mesure from the centimeter side!!
6 0
3 years ago
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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
nlexa [21]

Answer:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of 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}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{60}{6}= 10 and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0
2 years ago
The mass of the sun is 2.13525×1030 kilograms. The mass of Mercury is 3.285×1023 kilograms. How many times greater is the mass o
Helga [31]
<span>6.5 x 10^6 To answer this question, you need to divide the mass of the sun by the mass of mercury. So 2.13525 x 10^30 / 3.285 x 10^23 = ? To do the division, divide the mantissas in the normal fashion 2.13525 / 3.285 = 0.65 And subtract the exponents. 30 - 23 = 7 So you get 0.65 x 10^7 Unless the mantissa is zero, the mantissa must be greater than or equal to 0 and less than 10. So multiply the mantissa by 10 and then subtract 1 from the exponent, giving 6.5 x 10^6 So the sun is 6.5 x 10^6 times as massive as mercury.</span>
8 0
2 years ago
Read 2 more answers
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