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SCORPION-xisa [38]
3 years ago
5

Which relation is a function

Mathematics
1 answer:
Anon25 [30]3 years ago
3 0
\{(\boxed{2};\ 3);\ (1;\ 5);\ (\boxed{2};\ 7)\}-NO\\\\\{(-1;\ 5);\ (-2;\ 6);\ (-3;\ 7)\}-YES\\\\\{(\boxed{11};\ 9);\ (\boxed{11};\ 5);\ (9;\ 3)\}-NO\\\\\{(\boxed{3};\ 8);\ (0;\ 8);\ (\boxed{3};\ -2)\}-NO
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Solve system of linear equation with substitution
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Answer:

x=1, y=2

Step-by-step explanation:

solve by substitution so substitute x= 6y-11 in the next equation

3 (6y-11) - 2y = -1

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16y = 33-1

y= 32/16 =2

x= 6y -11 = 6*2 -11 = 12 -11 = 1

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3 years ago
A researcher wants to see if birds that build larger nests lay larger eggs. He selects two random samples of nests: one of small
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Answer:

95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.

Step-by-step explanation:

Confidence interval is given by mean +/- margin of error (E)

Eggs from small nest

Sample size (n1) = 60

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Sample variance = 24.7

Eggs from large nest

Sample size (n2) = 159

Mean = 35.6

Sample variance = 39

Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11

Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93

Difference in mean = 37.2 - 35.6 = 1.6

Degree of freedom = n1+n2 - 2 = 60+159-2 = 217

Confidence level = 95%

Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132

E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79

Lower limit = mean - E = 1.6 - 0.79 = 0.81

Upper limit = mean + E = 1.6 + 0.79 = 2.39

95% confidence interval for the difference in average mass is (0.81, 2.39)

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Step-by-step explanation:

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