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mario62 [17]
3 years ago
7

Which of the following is the third value for the data set below

Mathematics
1 answer:
Tcecarenko [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Ur mom

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Answer:

-17

Step-by-step explanation:

-6-11=-17

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How many 2 thirds cups servings are in 4 cup container of food
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There are 6 2/3 cup servings in a 4 cup container.
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2 years ago
Point b lies on segment AC. find the length of segment BC if AB = 4mm and AC = 4cm
Nataly_w [17]

Answer:

Segment BC = 36 mm or 3.6 cm

Step by Step Explanation:

There are 10 mm in 1 cm

meaning that segment AC is 40 mm long

if AB is 4 mm and it is one part of the two, then we need to subtract that from the total to find out what is left.

40 - 4 = 36

Segment BC = 36 mm or 3.6 cm

4 0
3 years ago
Find the mean, median, and mode of the following data set
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Answer: Mean- 17

Median- 15

Mode- 18

Step-by-step explanation: Your welcome. Branliest, please.

8 0
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:

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

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)}

4 0
3 years ago
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