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ZanzabumX [31]
2 years ago
7

Anna has been studying a type of bacteria that triples every month. Originally, there were 4 bacterial cells. She wants to know

how many there will be after 15 months. Which equation should she use?
A) a15 = 4(3)(15)
B) a15 = 4(3)15
C) a15 = 4(3)15 − 1
D) a15 = 4(3)(15 − 1)
Mathematics
1 answer:
bija089 [108]2 years ago
7 0

As per the given information;

Bacteria triples every month.

Originally, there were 4 bacterial cells.

We are suppose to find how many there will be after 15 months.

Hence we can say , the number of bacteria after n month can be given using the exponential growth function

a_n=4*3^n

So number of bacteria after 15 month will be

a_{15}=4*3^{15}

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x²<span>-20x
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the answer is
the number must be 100
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Por que es importante que los científicos desarrollen una forma de hacer los tejidos que se han construido en el sistema de sumi
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3 years ago
What is the explicit formula for the sequence?
Aleksandr [31]

Answer:

a_{n} = n - 2

Step-by-step explanation:

Note the difference between consecutive terms is constant, that is

0 - (- 1) = 0 + 1 = 1

1 - 0 = 1

2 - 1 = 1

3 - 2 = 1

This indicates the sequence is arithmetic with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = - 1 and d = 1, thus

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7 0
3 years ago
Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb
Maksim231197 [3]

Answer:

n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689

So the answer for this case would be n=689 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: "Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.

Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year."

Solution for the problem

First we need ti find the estimation for the standard deviation using the Rule of thumb, with the following formula:

s \approx \frac{R}{4}

Where R is the range defined as :

R = Max - Min = 8-0 = 8

So then the deviation would be approximately:

s \approx 4

Important concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error (ME) is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (1)

And on this case we have that ME =\pm 0.25 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (2)

We can assume that the estimator for the population deviation from the rule of thumb is \hat \sigma = s= 4

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got z_{\alpha/2}=1.64, replacing into formula (2) we got:

n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689

So the answer for this case would be n=689 rounded up to the nearest integer

6 0
3 years ago
A shopkeeper bought two watches for Rs. 400. He sold them of gain 5% on one and loss 5% of the other. What is his final gain or
nlexa [21]
This is short cut for this type of questions, ie, gain or loss square/100=5^2/100=>0.25%loss
Always this type of questions will have loss
6 0
3 years ago
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