A pond has a minnow population of 20000 that is increasing at a rate of 5%

1 answer:

7 0

20000x.05= 1000 added minnows per year, with the algae decline, we loose 750 minnows per year. 250 minnows are added every year. The rate 5% every year would be exponential growth, and the decline of 750 minnows per year is a result of a negative constant (not exponential decline.)

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Find the solution of y = 6x + 1 for x = 5

snow_lady [41]

Answer:

y equals to 31

Step-by-step explanation:

6(5) = 30

30+1 = 31

3 0

2 years ago

To illustrate the effects of driving under the influence (DUI) of alcohol, a police officer brought a DUI simulator to a local h

Oksanka [162]

Answer:

a) If we design the experiment on this way we can check if we have an improvement with the method used.

We assume that we have the same individual and we take a value before with the normal impaired condition and the final condition is the normal case.

b) $-0.96-2.306\frac{0.359}{\sqrt{9}}=-1.24$

$-0.96+2.306\frac{0.359}{\sqrt{9}}=-0.69$

The 95% confidence interval would be given by (-1.24;-0.69)

Step-by-step explanation:

Part a

If we design the experiment on this way we can check if we have an improvement with the method used.

We assume that we have the same individual and we take a value before with the normal impaired condition and the final condition is the normal case.

Part b

For this case first we need to find the differences like this :

Normal, Xi 4.47 4.24 4.58 4.65 4.31 4.80 4.55 5.00 4.79

Impaired, Yi 5.77 5.67 5.51 5.32 5.83 5.49 5.23 5.61 5.6

Let $d_i = Normal -Impaired$

$d_i : -1.3, -1.43, -0.93, -0.67,-1.52, -0.69, -0.68, -0.61, -0.81$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.96$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =0.359$

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=2.306$