All categories

2 years ago

Plz Hurry

Mathematics

1 answer:

Svetradugi [14.3K]2 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Find the greatest common factor of 50, 25, and 100

tatuchka [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

On the way to her babysitting job, Mary rode her bike for the first mile. She stopped at a traffic light and waited a few minute

Aloiza [94]

The answer is b, hope this helps

7 0

3 years ago

Read 2 more answers

During the exponential phase, E. coli bacteria in a culture increase in number at a rate proportional to the current population.

NikAS [45]

Answer:

197.2 million

Step-by-step explanation:

The appropriate exponential equation for the population is ...

p(t) = 172.0e^(0.019t)

Then we can compute for t=7.2:

p(7.2) = 172.0e^(0.019·7.2) ≈ 172.0·1.146599 ≈ 197.2

7.2 minutes from now, the population will be about 197.2 million.

___

For continuous growth (or continuous compounding), the exponential formula is ...

f(t) = (value at t=0)×e^(rt)

where r is the growth rate in one unit of time, and t is the number of time periods.

8 0

3 years ago

What’s the midpoint P(0,-2), Q(1,1

LiRa [457]

You would have to add x1 + x2 and divide by 2. do the same with y1 and y2.

(1/2,-1/2)

6 0

2 years ago

Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

$=P(0.8)^{2}$ -------------------------(2)

Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

3 0

3 years ago