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3 years ago

13

A cafeteria worker who fails to wash his hands thoroughly and fails to wear gloves inoculates a quiche with 4 E. coli when he us

es his finger to test whether it is done. By the time you purchase the quiche, there are 128 E. coli cells in it. How many generations did the cells go through?

1 answer:

marysya [2.9K]3 years ago

7 0

Answer:

5 generations

Step-by-step explanation:

Data provided in the question:

Number of bacteria at the beginning of a time interval , B = 4

Number of bacteria at the end of the time interval, b = 128

Now,

The relation between the bacteria at the beginning, bacteria at the end and the number of generations (n) is given as:

b = B × 2ⁿ

Therefore,

on substituting the respective values, we get

128 = 4 × 2ⁿ

or

2ⁿ = 32

or

2ⁿ = 2⁵

or

n = 5

hence,

5 generations

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Read 2 more answers

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valentinak56 [21]

Answer:

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$4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39$

The 95% confidence interval is given by (3.61;4.39)

Step-by-step explanation:

Notation and definitions

n=15 represent the sample size

$\bar X=4$ represent the sample mean

$s=0.7$ represent the sample standard deviation

m represent the margin of error

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".

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n-1=15-1=14$

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,14)" for $t_{\alpha/2}=-2.14$

"=T.INV(1-0.025,14)" for $t_{1-\alpha/2}=2.14$

The critical value $tc=\pm 2.14$

Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=2.14 \frac{0.7}{\sqrt{15}}=0.387$

Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61$

$4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39$

The 95% confidence interval is given by (3.61;4.39)

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3 years ago

If sin theta = 2/3 and sex theta < 0 , find cos theta and tan theta

FromTheMoon [43]

Answer:

$\displaystyle cos\theta=-\frac{\sqrt{5}}{3}$

$\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

<u>Trigonometric Formulas</u>

To solve this problem, we must recall some basic relations and concepts.

The main trigonometric identity relates the sine to the cosine:

$sin^2\theta+cos^2\theta=1$

The tangent can be found by

$\displaystyle tan\theta=\frac{sin\theta}{cos\theta}$

The cosine and the secant are related by

$\displaystyle cos\theta=\frac{1}{sec\theta}$

They both have the same sign.

The sine is positive in the first and second quadrants, the cosine is positive in the first and fourth quadrants.

The sine is negative in the third and fourth quadrants, the cosine is negative in the second and third quadrants.

We are given

$\displaystyle sin\theta=\frac{2}{3}$

Find the cosine by solving

$sin^2\theta+cos^2\theta=1$

$\displaystyle \left(\frac{2}{3}\right)^2+cos^2\theta=1$

$\displaystyle cos^2\theta=1-\left(\frac{2}{3}\right)^2=1-\frac{4}{9}=\frac{5}{9}$

$\displaystyle cos\theta=\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}$

$\boxed{\displaystyle cos\theta=-\frac{\sqrt{5}}{3}}$

We have placed the negative sign because we know the secant ('sex') is negative and they both have the same sign.

Now compute the tangent

$\displaystyle tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}$

Rationalizing

$\displaystyle tan\theta=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}$

$\boxed{\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}}$

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