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e-lub [12.9K]
3 years ago
9

A number cube with faces labeled from 1 to 6 was rolled 20 times. Each time the number

Mathematics
1 answer:
aliina [53]3 years ago
3 0
3+5=8
8/20 =.4
.4 goes to 40% bc you move the decimal over twice
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Answer:

11

Step-by-step explanation:

divide 209 by 19 to get an equal number of games each month (for 19 months), you will get 11 games each month.

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3 years ago
Create an equation with at least two grouping symbols for which<br> there is no solution
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5+7-4=6
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3 years ago
The distances of a port from 20 differ- ent locations form an AP. If the farthest distance is 300 km, and the nearest distance i
Mkey [24]

Answer:

Required difference = 270/19 km

Step-by-step explanation:

If we consider the AP of distances to be

a, a + r, a + 2r, ..., a + 19r,

where a is the distance of the nearest location from the port and r is the difference between any two successive location.

Given that,

a + 19r = 300 .....(1)

a = 30 ..... (2)

Using (2), from (1), we get

30 + 19r = 300

or, 19r = 300 - 30

or, 19r = 270

or, r = 270/19

Therefore the distance between any two successive location is 270/19 km.

5 0
3 years ago
Uninhibited growth can be modeled by exponential functions other than​ A(t) ​=Upper A 0 e Superscript kt. For ​ example, if an i
laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

3 0
3 years ago
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