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vagabundo [1.1K]
2 years ago
12

Harriet is cultivating a strain of bacteria in a petri dish. Currently, she has 103 bacteria in the dish. The bacteria divide ev

ery two hours such that the number of bacteria has doubled by the end of every second hour. How many bacteria will Harriet have in the dish at the end of 6 hours
Mathematics
2 answers:
SVEN [57.7K]2 years ago
8 0
Based on the conditions given above, the number of bacteria at any time t (in hours) is calculated by the equation,
                                         at = (a1)(2^t/2)

where a1 is the initial number of bacteria and at is the number at any time t. Substituting the givens,
                                      a6 = (103)(2^6/2) = 824 

Thus, there are 824 bacteria after 6 hours. 
fomenos2 years ago
6 0

Formula for growth or decay

R= R_{0}[1 \pm \frac{r}{100}]^t

R= Final population of Bacteria

R_{0}= Initial Population

r = Rate of growth or decay

t= Time period

R_{0}= 103 bacteria

when, t=2

R becomes 103 × 2= 206 bacteria

Substituting these values in the formula of growth

206 = 103 × (1+ \frac{r}{100})^2

Dividing both sides by 103, we get

2 = (1+ \frac{r}{100})^2

(1+ \frac{r}{100})= √2------(1)

(1+ \frac{r}{100})= 1.414

r = (1.414 -1) × 100= .414 × 100= 41.4%

Number of bacteria at the end of 6 hours = 103 × (1+ \frac{r}{100})^6= 103 × [\sqrt{2}]^6 ------Using (1)

= 103 × 2³

=103 × 8

= 824→→bacteria that Harriet have in the dish at the end of 6 hours


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Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

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