Solve 2(m-3) equal -6

Let the abbreviation PSLT stand for the percent of the gross family income that goes into paying state and local taxes. Suppose

gavmur [86]

Answer:

Number of families that should be surveyed if one wants to be 90% sure of being able to estimate the true mean PSLT within 0.5 is at least 43.

Step-by-step explanation:

We are given that one wants to estimate the mean PSLT for the population of all families in New York City with gross incomes in the range $35.000 to $40.000.

If sigma equals 2.0, we have to find that how many families should be surveyed if one wants to be 90% sure of being able to estimate the true mean PSLT within 0.5.

Here, we will use the concept of Margin of error as the statement "true mean PSLT within 0.5" represents the margin of error we want.

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<u>SO, Margin of error formula is given by;</u>

Margin of error = $Z_(_\frac{\alpha}{2}_ ) \times \frac{\sigma}{\sqrt{n} }$

where, $\alpha$ = significance level = 10%

$\sigma$ = standard deviation = 2.0

n = number of families

Now, in the z table the critical value of x at 5% ( $\frac{0.10}{2} = 0.05$ ) level of significance is 1.645.

SO, Margin of error = $Z_(_\frac{\alpha}{2}_ ) \times \frac{\sigma}{\sqrt{n} }$

0.5 = $1.645 \times \frac{2}{\sqrt{n} }$

$\sqrt{n} =\frac{2\times 1.645 }{0.5}$

$\sqrt{n} =6.58$

n = $6.58^{2}$

= 43.3 ≈ 43

Therefore, number of families that should be surveyed if one wants to be 90% sure of being able to estimate the true mean PSLT within 0.5 is at least 43.

4 0

4 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: