What is the justification for step 3 in the solution process?

How many hours will it take a culture of bacteria to increase from 20 to 2000 if the growth rate per hour is 85%

galben [10]

Answer:

Approximately

7.5

hours

Step-by-step explanation:

Explanation:

We are trying to determine the value of

h

in the compound growth formula

XXX

20

×

(

1

+

0.85

)

h

=

2000

or, simplified:

XXX

1.85

h

=

100

Taking the

log

1.85

of both sides:

XXX

log

1.85

100

With the aid of my trusty spreadsheet:

XXX

h

≈

7.48582192

3 0

3 years ago

Whats a prime number

skelet666 [1.2K]

24 maybe because 24 is a even number so it has to be 24 as the answer

3 0

3 years ago

Read 2 more answers

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

What are all of the keys that must be pressed, in correct

Marta_Voda [28]

The answer is number 1

5 0

3 years ago

Read 2 more answers

Andrew adds two numbers. Each addend has seven digits. The sum has eight digits. What digit is in the ten millions place in the

tatuchka [14]

Answer:

5,000,000 and 6,000,000

Step-by-step explanation:

5,000,000 and 6,000,000 is 11,000,000

or 5,000,000 and 5,000,000 is 10,000,000

and the 1 is in the ten millions place in the sum

4 0

2 years ago