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2 years ago

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plz hurryA 3-column table with 6 rows. Column 1 is labeled Number Rolled with entries 1, 2, 3, 4, 5, 6. Column 2 is labeled Obse

rved Frequency with entries 10, 12, 10, 10, 8, 10. Column 3 is labeled Relative frequency with entries StartFraction 1 over 6 EndFraction, A, StartFraction 1 over 6 EndFraction, StartFraction 1 over 6 EndFraction, B, StartFraction 1 over 6 EndFraction. A number cube is rolled 60 times. The results of those 60 trials are recorded in the table. Complete the table. A = B =

1 answer:

olga55 [171]2 years ago

4 0

Answer:

The complete table is shown below.

Step-by-step explanation:

The data provided is:

Number Rolled Observed Frequency Relative frequency

1 10 $\frac{10}{60}=\frac{1}{6}$

2 12 $\frac{12}{60}=\frac{1}{5}$

3 10 $\frac{10}{60}=\frac{1}{6}$

4 10 $\frac{10}{60}=\frac{1}{6}$

5 8 $\frac{8}{60}=\frac{2}{15}$

6 10 $\frac{10}{60}=\frac{1}{6}$

TOTAL 60 1

The complete table is shown above.

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Step-by-step explanation:

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The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

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Answer:

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2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

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0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

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Answer:

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Step-by-step explanation:

The slope-intercept form of an equation of a line:

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