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monitta
4 years ago
12

A bacterium cell splits into 2 cells every hour. Write and evaluate and exponenetial expression to find how many cells there wil

l be in 6 hours. Then use your answer to help you find the number of hours it will take for there to be 1,024 cells?..
Mathematics
1 answer:
viktelen [127]4 years ago
5 0

Answer:

There will be 64 cells in 6 hours. It will take 10 hours for there to be 1,024 cells.

Step-by-step explanation:

For this question, we need to think exponents. When the cell splits, two cells arise, then 4, then 8, and so on. Each time, the count doubles, so 2 will be the base, with 6 being the exponent due to 6 hours having elapsed.

Expression: 2^6=x\\64=x

Using our equation will be a lot more useful for the second part of the question. We'll still use 2 as the base, but we're using x as the exponent (x represents the amount of hours it takes to reach y cells, which in this problem is 1,024):

2^{x} =1,024\\log_{2}1,024=x\\ 10=x\\or\\2^x=2^10\\x=10

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Yes because -5 does equal -5
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3 years ago
The nth term of a sequence is 20-n^2
Nimfa-mama [501]

Answer:

a) 20 - 3^2 = 11

b) 5th, 20 - 5^2 = -5

Step-by-step explanation:

The answer assumes your terms begin at 1. If the terms begin at zero

a) 16

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if you chose to be extremely technical you would take the square root of 20.1 or 20.00000009 but I'm assuming they want whole numbers

8 0
3 years ago
Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply,
myrzilka [38]

Answer:

163.83-4.03\frac{20.094}{\sqrt{6}}=130.77    

163.83+4.03\frac{20.094}{\sqrt{6}}=196.89    

So on this case the 99% confidence interval would be given by (130.77;196.89)    

Step-by-step explanation:

Previous 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 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".

Data: 175 177 175 180 138 138

We can calculate the mean and the deviation from these data with the following formulas:

\bar X= \frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=163.83 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=20.093 represent the sample standard deviation

n=6 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=6-1=5

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,5)".And we see that t_{\alpha/2}=4.03

Now we have everything in order to replace into formula (1):

163.83-4.03\frac{20.094}{\sqrt{6}}=130.77    

163.83+4.03\frac{20.094}{\sqrt{6}}=196.89    

So on this case the 99% confidence interval would be given by (130.77;196.89)    

7 0
3 years ago
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For the answer to the question above,
 <span>r = 1 + cos θ 

x = r cos θ 
x = ( 1 + cos θ) cos θ 
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y = r sin θ 
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dy/dx = (dy/dθ) / (dx/dθ) 
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For horizontal tangent line, dy/dθ = 0 

cos θ - sin^2 θ + cos^2 θ = 0 
cos θ - (1-cos^2 θ) + cos^2 θ = 0 
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2y^2+y-1=0 
2y^2+2y-y-1=0 
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(y+1)(2y-1)=0 
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cos θ =-1 
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θ = π/3 , π, 5π/3 
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</span>---------------------------------------------------------------------------------
For horizontal tangent line, dx/dθ = 0 

<span>-sin θ - 2 cos θ sin θ = 0 </span>
<span>-sin θ (1+ 2 cos θ ) = 0 </span>
<span>sin θ = 0 </span>
<span>θ = 0, π </span>

<span>(1+ 2 cos θ ) =0 </span>
<span>cos θ =-1/2 </span>
<span>θ = 2π/3 </span>
<span>θ = 4π/3 </span>

<span>θ = 0, 2π/3 ,π, 4π/3 </span>
<span>when θ = 0, r=2 </span>
<span>when θ = 2π/3, r=1/2 </span>
<span>when θ = π, r=0 </span>
<span>when θ = 4π/3 , r=1/2 </span>

<span>(2,0) , (1/2, 2π/3) , (0, π), (1/2, 4π/3) </span>
<span>At (2,0) there is a vertical tangent line</span>
7 0
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Sonbull [250]
I believe I would be c but I could be wrong
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