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wel
3 years ago
9

brett has been studying a type of bacteria that doubles every month. Originally, there were 5 bacterial cells. He wants to know

how many there will be after 42 months?
Mathematics
2 answers:
Sedbober [7]3 years ago
8 0

Answer:

Step-by-step explanation:

Originally, there were only five bacterial cells. After one month, the amount of bacteria is doubled with ten bacteria. The growth is represented by the formula:

a42 = 5 x 2^1

After 42 months, the growth can be solved using this formula:

a42 = 5 x 2^42 just so you know my cuz help me shes good in this

igor_vitrenko [27]3 years ago
4 0

Answer:

N=2.1990232556 \times 10^{13}

Step-by-step explanation:

Given : Brett has been studying a type of bacteria that doubles every month. Originally, there were 5 bacterial cells.

To Find: He wants to know how many there will be after 42 months?

Solution:

Since we are given that initially there were 5 bacterial cells.

Bacteria doubles every month

Let n denotes the number of months .

Function becomes : N=N_0(2)^n

N_0 = initial amount

N = amount after n months

So, N=5(2)^n

Substitute n = 42

N=5(2)^{42}

N=2.1990232556 \times 10^{13}

Thus there will be 2.1990232556 \times 10^{13} bacteria after 42 months.

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Two separate tests are designed to measure a student's ability to solve problems. Several students are randomly selected to take
Ipatiy [6.2K]

Answer:

a) r = 0.974

b) Critical value = 0.602

Step-by-step explanation:

Given - Two separate tests are designed to measure a student's ability to solve problems. Several students are randomly selected to take both test and the results are give below

Test A | 64 48 51 59 60 43 41 42 35 50 45

Test B |  91 68 80 92 91 67 65 67 56 78 71

To find - (a) What is the value of the linear coefficient r ?

             (b) Assuming a 0.05 level of significance, what is the critical value ?

Proof -

A)

r = 0.974

B)

Critical Values for the Correlation Coefficient

n       alpha = .05          alpha = .01

4           0.95                       0.99

5           0.878                     0.959

6           0.811                       0.917

7           0.754                      0.875

8           0.707                      0.834

9           0.666                      0.798

10          0.632                      0.765

11           0.602                      0.735

12          0.576                       0.708

13          0.553                       0.684

14           0.532                       0.661

So,

Critical r = 0.602 for n = 11 and alpha = 0.05

6 0
3 years ago
The answer to this question
Sophie [7]

The answer to this question is D) 13.9 in.

Try using the Pythagorean theorem when solving to get the hypotenuse or other angles for a right triangle.

4 0
3 years ago
I am struggling to figure out the last steps of this question:
igor_vitrenko [27]
I have attached a chart of the given information. Using subtraction from 100 and the other totals, you should be able to figure out the answer. If not, comment and I will send the completed chart

8 0
3 years ago
A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's hal
mars1129 [50]

Answer:

  t=5.0days

Step-by-step explanation:

Using the formula for the exponential decay that is N=N_{0}e^{-kt}, we have N=\frac{1}{2}{\times}46=23, N_{0}=46 and k=0.1374.

Thus, N=N_{0}e^{-kt} becomes

23=46{\times}e^{-0.1374t}

\frac{23}{46}=e^{-0.1374t}

\frac{1}{2}=e^{-0.1374t}

Taking log on both sides, we get

ln(\frac{1}{2})={-0.1374t}

t=\frac{ln\frac{1}{2}}{-0.1}

t=\frac{-0.6931}{-0.1374}

t=5.0days

4 0
3 years ago
HIGH POINTS/WILL MARK BRAINLIEST
Art [367]

Answer:

The leading term is 2x^7. Since n is odd and a is positive, the end behavior is down and up.

5 0
2 years ago
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