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3 years ago

10

Evaluate the expression for the given variable.

1 answer:

Reika [66]3 years ago

7 0

Answer:

8

Step-by-step explanation:

Substitute x = 3 into the expression

= $\frac{(9-3)^2+4}{5}$

= $\frac{6^2+4}{5}$

= $\frac{36+4}{5}$

= $\frac{40}{5}$

= 8

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Sara tried to evaluate 21 : 7 x 8 step-by-step

nlexa [21]

Answer: Sara multiplied by 3 when she should have multiplied by 8.

Step-by-step explanation:

21 : 7*8

7*8 = 56

21 : 56

The simplest form of this ratio is 3 : 8

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3 years ago

Suppose that the hourly wages of fast food workers are normally distributed with an unknown mean and standard deviation. The wag

Nutka1998 [239]

Answer: 1.303639

Step-by-step explanation:

The t-score for a level of confidence $(1-\alpha)$ is given by :_

$t_{(df,\alpha/2)}$ , where df is the degree of freedom and $\alpha$ is the significance level.

Given : Level of significance : $1-\alpha:0.80$

Then , significance level : $\alpha: 1-0.80=0.20$

Sample size : $n=40$

Then , the degree of freedom for t-distribution: $df=n-1=40-1=39$

Using the normal t-distribution table, we have

$t_{(df,\alpha/2)}=t_{39,0.10}=1.303639$

Thus, the t-score should be used to find the 80% confidence interval for the population mean =1.303639

8 0

3 years ago

Need help ASAP. Question in photo. Will mark brainliest if correct

ioda

Answer:

C

Step-by-step explanation:

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2 years ago

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yan [13]

Answer: 2n + 1 = n + 11

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3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago