The population of the town in 1960 is 48.80 thousands
<h3>How to determine the population in 1950?</h3>
The equation of the model is given as:
f(t) = 42e^(0.015t)
1960 is 10 years after 1950.
This means that:
t = 10
Substitute t = 10 in f(t) = 42e^(0.015t)
f(10) = 42e^(0.015 * 10)
Evaluate
f(10) = 48.80
Hence, the population of the town in 1960 is 48.80 thousands
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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
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)
Answer:
Total amount of fencing needed as an algebraic expression in terms of x is: <em>10x</em><em> </em><em>+</em><em> </em><em>3</em> .
Step-by-step explanation:
As it is given that each rectangle has the same dimensions, the dimensions of each rectangle must be: x units by 2x + 1 units.
Based on this, we can calculate the total amount of fencing needed.
Let width of each rectangle = x
Let length of each rectangle = 2x + 1
There are 4 widths and 3 lengths in total of fencing.
Therefore:
= 4 ( x ) + 3 ( 2x + 1 )
Expand:
= 4x + 6x + 3
Group like-terms:
= 10x + 3
Answer:
x^(2)-3x-16=7x
Step-by-step explanation: