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ohaa [14]
3 years ago
7

What is the median for the following scores for Mrs. Johnson's first test? 59, 22, 87, 44, 96, 38, 70, 85, 79

Mathematics
2 answers:
katrin2010 [14]3 years ago
8 0

To work out the median, you need to numerically sort the data (because currently, it's unordered).

The ordered set of data is: 22, 38, 44, 59, 70, 79, 85, 87, 96

Now you just need to pick the middle value, which is 70:

(as you can see: 22, 38, 44, 59, 70, 79, 85, 87, 96)

Therefore, your answer is 70 :)

Bess [88]3 years ago
3 0

Hello,

59, 22, 87, 44, 96, 38, 70, 85, 79

=> 22 ; 38 ; 44 ; 59 ; 70 ; 79 ; 85 ; 87 ; 96

The median is 70.

Bye :)

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72

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

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In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

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