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

Mean = 37.2

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)

3 0

2 years ago

Mrs. Levi's class raised $217.35 in one-time donations for the library's new book collection. The class also had 4 people pledge

nexus9112 [7]

Answer: $817.35

Step-by-step explanation:

150x4 = 600

600+217.35 = 817.35

7 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Mashcka [7]

Multiplying eqn no. two with 4 we add it up with eqn 1.

$4x - 5y = 10 \\ - 4x - 12y = 32 \\$

$- 17y = 42 \\ y = \frac{ - 42}{17}$

When

$- x - 8 = 3y \\ x + 3y = - 8$

$x = - 8 - 3 \frac{ - 42}{17 } \\ x = - 8 + \frac{42}{17} \\ x = \frac{ - 10}{17}$

So, x = - 10/17 and y = - 42/17.

5 0

3 years ago

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