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Furkat [3]
3 years ago
15

PLEASE HELP!!

Mathematics
1 answer:
ruslelena [56]3 years ago
3 0

Answer: 104 miles

Step-by-step explanation:

If 2.5 in = 52 mi, and 2.5 × 2 = 5, then 2.5(2) = 52(2)

5 in = 104 mi

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irina1246 [14]

Based on the table (see attachment), the time which corresponds to the mode of this data set is: B. 9:00 P.M.

<h3>What is mode?</h3>

A mode simply refers to a statistical term that is used to denote the value that appears most often or occurs repeatedly in a given data set.

This ultimately implies that, a mode represents the value (number) with the highest frequency and this is 9:00 P.M with a frequency of 25.

Read more on mode here: brainly.com/question/542771

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6 0
2 years ago
Please factorise 16p²-9q² for me<br>​
kirill115 [55]

Answer:

a ^ 2 - b ^ 2 = ( a + b ) ( a - b ) where a = 4p and b = 3q.

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A man 50 years old has 8 sons born of equal intervals. The sum of the ages of the father and sons is 186. What is the age of the
Rudiy27
<span>let x be the interval, then: 186 = 50 + 3 + (3+x) + (3+2x) + (3+3x) + (3+4x) + (3+5x) + (3+6x) + (3+7x) 186 = 74 + 28x x = 4 Eldest son age = 3+7x = 3+28 = 31.</span>
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- Which expression shows 50 + 30 written as a product of two factors?
asambeis [7]

Option 4 is the answer.

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Hope it helps!

7 0
3 years ago
Read 2 more answers
A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria
Kaylis [27]

Answer:

A) The expression for the number of bacteria is P(t) = 400e^{0.7783t}.

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

P' = kP

where P stands for the population of bacteria.

Writing P' as \frac{dP}{dt}, we get

\frac{dP}{dt} = kP.

Notice that this is a separable equation, so

\frac{dP}{P} = kdt.

Then, integrating in both sides of the equality:

\int\frac{dP}{P} = \int kdt.

We have,

\ln P = kt+C.

Now, taking exponential

P(t) = Ce^{kt}.

The next step is to find the value for the constant C. We do this using the initial condition P(0)=400. Recall that this is the initial population of bacteria. So,

400 = P(0) = Ce^{k0}=C.

Hence, the expression becomes

P(t) = 400e^{kt}.

Now, we find the value for k. We are going to use that P(4)=9000. Notice that

9000 = 400e^{k4}.

Then,

\frac{90}{4} = e^{4k}.

Taking logarithm

\ln\frac{90}{4} = 4k, so \frac{1}{4}\ln\frac{90}{4} = k.

So, k=0.7783788273, and approximating to the fourth decimal place we can take k=0.7783. Hence,

P(t) = 400e^{0.7783t}.

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

P(5) =400e^{0.7783*5} = 19593.723 \approx 19593.  

C) In this case we want to do the reverse operation: we want to find the value of t such that

30000 = 400e^{0.7783t}.

This expression is equivalent to

75 = e^{0.7783t}.

Now, taking logarithm we have

\ln 75 = 0.7783t.

Finally,

t = \frac{\ln 75}{0.7783} \approx 5.55.

So, after 5.55 hours the population of bacteria will reach 30000.

6 0
3 years ago
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