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3 years ago

Help me solve this problem!!

Mathematics

1 answer:

Whitepunk [10]3 years ago

4 0

Answer:

initial size: 75
doubling time: 7.51 minutes
after 115 minutes: about 3,056,900
reaches 11,000: 54.03 minutes

Step-by-step explanation:

For given points (t1, y1), (t2, y2), I like to write the exponential function as ...

y(t) = y1·(y2/y1)^((t-t1)/(t2-t1))

This can be converted to other forms (such as a·b^t or a·e^(kt)) fairly easily, but those tend not to reproduce the given numbers exactly as this form does.

Using (15, 300) and (35, 1900) as our data values, the exponential function can be written as ...

y(t) = 300·(19/3)^((t-15)/20)

a) The initial size of the culture is the value of y(0).

y(0) = 300·(19/3)^(-15/20) ≈ 75.144

y(0) ≈ 75 . . . initial population

b) The doubling period will be the value of t that satisfies ...

(19/3)^(t/20) = 2

Taking logarithms, we have ...

(t/20)·log(19/3) = log(2)

t = 20·log(2)/log(19/3) ≈ 7.5104 . . . . minutes

The doubling time is about 7.51 minutes.

c) Evaluating the formula for t=115, we have ...

y(115) = 300·(19/3)^(100/20) ≈ 3056912.346

The count after 115 minutes will be about 3,056,900.

d) Solving y(t) = 11,000, we have ...

11000 = 300·(19/3)^((t-15)/20)

11000/300 = (19/3)^((t-15)/20)

log(110/3) = (t-15)/20·log(19/3)

t = 20·log(110/3)/log(19/3) + 15 ≈ 54.027

It will take about 54.03 minutes for the count to reach 11,000.

_____

I find a graphing calculator to be a nice tool for solving problems like this.

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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

First find the slope of the line

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