Answer:
y = 6x - 11
Step-by-step explanation:
An arithmetic sequence is a linear function. This means it steadily increases or decreases through addition/subtraction by a constant amount. Here the sequence grows by adding 6 each time.
-11 + 6 = -5
-5 + 6 = 1
1 + 6 = 7
etc....
Since this is linear, its equation has the form y = mx + b where m is the slope and b is the y-intercept. The slope is m = 6 and b = -11.
y = 6x - 11
Answer:
a) 151lb.
b) 6.25 lb
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, a large sample size can be approximated to a normal distribution with mean and standard deviation 
.
In this problem, we have that:
So
a) The expected value of the sample mean of the weights is 151 lb.
(b) What is the standard deviation of the sampling distribution of the sample mean weight?
This is 
The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.
The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.
This leaves us with the second choice. Recall that
1 + tan²(x) = sec²(x)
and the derivative of tangent,
(tan(x))' = sec²(x)
Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then
y' = sec²(x³/3 + 2) • x²
and substituting y and y' into the ODE gives
sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))
x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)
which is an identity.
So the solution is y = tan(x³/3 + 2).
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