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choli [55]
3 years ago
5

A population of bacteria is treated with an antibiotic. It is estimated that 5,000 live bacteria existed in the sample before tr

eatment. After each day of treatment, 40% of the sample remains alive. Which best describes the graph of the function that represents the number of live bacteria after x days of treatment?
f(x) = 5000(0.4)x, with a horizontal asymptote of y = 0
f(x) = 5000(0.6)x, with a vertical asymptote of x = 0
f(x) = 5000(1.4)x, with a horizontal asymptote of y = 0
f(x) = 5000(1.6)x, with a vertical asymptote of x = 0
Mathematics
1 answer:
kodGreya [7K]3 years ago
7 0

Answer:

Step-by-step explanation:

The function is modeled after

y=a(b)^x where a is the initial amount and b is the growth/decay factor. If we start with 5000 bacteria, then a = 5000. The decay factor is determined by the rate at which the bacteria decrease. We know that rate at which the bacteria decrease is 60% because 40% are left alive; therefore, b = .4:

f(x)=5000(.4)^x with a horizontal asymptote of y = 0, the first choice.

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Tamiku [17]

Answer:

log7

Step-by-step explanation:

when adding logs, apply the log rule: \log _a\left(x\right)+\log _a\left(y\right)=\log _a\left(xy\right)

∴  \log\left(\frac{14}{3}\right)+\log\left(\frac{11}{5}\right)=\log\left(\frac{14}{3}\cdot \frac{11}{5}\right)

when subtracting logs, apply the log rule: \log _a\left(x\right)\:-\:\log _a\left(y\right)=\log _a\left(\frac{x}{y}\right)

\log\left(\frac{14}{3}\cdot \frac{11}{5}\right)-\log\left(\frac{22}{15}\right)=\log\left(\frac{\frac{14}{3}\cdot \frac{11}{5}}{\frac{22}{15}}\right)\\\\=\log\left(7\right)

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2 years ago
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Lina20 [59]
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Suppose that Y1, Y2,..., Yn denote a random sample of size n from a Poisson distribution with mean λ. Consider λˆ 1 = (Y1 + Y2)/
Burka [1]

Answer:

The answer is "\bold{\frac{2}{n}}".

Step-by-step explanation:

considering Y_1, Y_2,........, Y_n signify a random Poisson distribution of the sample size of n which means is λ.

E(Y_i)= \lambda \ \ \ \ \ and  \ \ \ \ \ Var(Y_i)= \lambda

Let assume that,  

\hat \lambda_i = \frac{Y_1+Y_2}{2}

multiply the above value by Var on both sides:

Var (\hat \lambda_1 )= Var(\frac{Y_1+Y_2}{2} )

            =\frac{1}{4}(Var (Y_1)+Var (Y_2))\\\\=\frac{1}{4}(\lambda+\lambda)\\\\=\frac{1}{4}( 2\lambda)\\\\=\frac{\lambda}{2}\\

now consider \hat \lambda_2 = \bar Y

Var (\hat \lambda_2 )= Var(\bar Y )

             =Var \{ \frac{\sum Y_i}{n}\}

             =\frac{1}{n^2}\{\sum_{i}^{}Var(Y_i)\}\\\\=\frac{1}{n^2}\{ n \lambda \}\\\\=\frac{\lambda }{n}\\

For calculating the efficiency divides the \hat \lambda_1 \ \ \ and \ \ \ \hat \lambda_2 value:

Formula:

\bold{Efficiency = \frac{Var(\lambda_2)}{Var(\lambda_1)}}

                  =\frac{\frac{\lambda}{n}}{\frac{\lambda}{2}}\\\\= \frac{\lambda}{n} \times \frac {2} {\lambda}\\\\ \boxed{= \frac{2}{n}}

8 0
3 years ago
Given equations A and B as 2/5x+y=12 and 5/2y-x=6 respectively which expression will eliminate the variable X a) 5/2A+2/5B b)5/2
frutty [35]

Answer:

C) A+\frac{2}{5}B

Step-by-step explanation:

Given Equations:

A) \frac{2}{5}x+y=12

B) \frac{5}{2}y-x=6

In order to eliminate x we need to multiply \frac{2}{5} to equation B and add it to A.

Multiplying \frac{2}{5} to B.

⇒ \frac{2}{5}\times (\frac{5}{2}y-x)=\frac{2}{5}\times 6

⇒ (\frac{2}{5}\times\frac{5}{2}y)-(\frac{2}{5}x)=\frac{12}{5}

⇒ y-\frac{2}{5}x=\frac{12}{5}

Adding it equation A.

   \frac{2}{5}x+y=12

+ -\frac{2}{5}x+y=\frac{12}{5}

We get 2y=\frac{72}{5}

Thus x is eliminated.

So, the expression used to eliminate x is:

A+\frac{2}{5}B

4 0
3 years ago
(1 point) The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with a mean of 12
garik1379 [7]

Answer:

a) Z = -2.7

b) Z = 2.3

c) Kyle's systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.

d) X = 144.5

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 127, \sigma = 10

(a) 100 millimeters z-score =

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 127}{10}

Z = -2.7

(b) 150 millimeters z-score =

Z = \frac{X - \mu}{\sigma}

Z = \frac{150 - 127}{10}

Z = 2.3

(c) Kyle's doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of his score?

1.75 standard deviations above the mean systolic blood pressure of males.

The problem does not talk about age.

So the answer is:

Kyle's systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.

d) Calculate Kyle's blood pressure

This is X when Z = 1.75. So

Z = \frac{X - \mu}{\sigma}

1.75 = \frac{X - 127}{10}

X - 127 = 1.75*10

X = 144.5

7 0
4 years ago
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