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Lostsunrise [7]
2 years ago
14

A and b are natural numbers. The diameter of cell a is 2.3 , and the diameter of cell b is 2.3 .

Mathematics
2 answers:
Pani-rosa [81]2 years ago
8 0

After clarification we see the respective diameters are

A = 2.3 \times 10^{-a}

B = 2.3 \times 10^{-b}

We're asked about the ratio

A:B =  2.3 \times 10^{-a} :  2.3 \times 10^{-b} = 10^{-a} : 10^{-b} = 10^b : 10^a = 10^{b-a} : 1

Now we can go through the choices.

A. If a = 4 and b = 6, then cell a is 100 times greater than cell b.

A:B=10^{6-4}:1 = 100:1

It would be more proper to say the diameter of cell A is 100 times the diameter of cell B, but yes, this is TRUE.

B. If a = 4 and b = 6, then cell b is 100 times greater than cell a.

FALSE. 10 to the minus 4 is 100 times bigger than 10 to the minus 6, so a is bigger.

C. If a = 4 and b = 6, then cell a is 2 times greater than cell b.

FALSE. We already calculated its 100 times bigger.

D. If a = 4 and b = 6, then cell b is 2 times greater than cell a.

FALSE.


Svetlanka [38]2 years ago
8 0

Answer:

A. If a = 4 and b = 6, then cell a is 100 times greater than cell b.

Step-by-step explanation:

2.3/2.3= 1

10(-4)/10(-6)=10(2)

your answer is 100, cell a is 100 times greater than cell b.

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Type two statements that use nextint() to print 2 random integers between (and including) 100 and 149. end with a newline. ex: 1
solong [7]

NB- Solution is emboldened

 

import java.util.Scanner;

import java.util.Random;

public class RandomGenerateNumbers {

public static void main (String [] args) {

Random randGen = new Random();

int seedVal = 0;

seedVal = 4;

randGen.setSeed(seedVal);

System.out.println(randGen.nextInt(50) + 100);

System.out.println(randGen.nextInt(50) + 100);

return;

}

}

4 0
1 year ago
Read 2 more answers
If you diluted 300 mL of 8% benzocaine lotion to 5% how much could you produce
Nataliya [291]

Answer:

you are diluting it 8/5 times, or 1.6 times.

You could make 230*1.6 ml of 5 percent lotion.

4 0
2 years ago
Cindy found a collection of baseball cards in her attic worth $8,000. the collection is estimated to increase in value by 1.5% p
Sergio [31]
A]
Exponential function is given by the form:
y=a(b)ˣ
where:
a=initial value
b=growth factor
From the question:
a=$8000, b=1.015, 
thus the exponential growth function of this situation is:
y=8000(1.015)ˣ

b] The value of the collection after 7 years will be:
x=7 years
Using the formula:
y=8000(1.015)ˣ
plugging the values we get:
y=8000(1.015)⁷
y=8,878.76

Answer: $8,878.76
7 0
2 years ago
A pair of fair dice is cast. what is the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one
WITCHER [35]
If a pair of fair dice is cast, the probability that the <span>sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is given by:

\frac{P(sum\, of\, 9\, \cup\,3)}{P(3)}

The number of outcomes of sum of 9 where at last one is 3 is (3, 6) and (6, 3) = 2.

</span>The number of outcomes of last one of the numbers falling uppermost is a 3 is (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3) and (6, 3) = 11.

Therefore, <span>the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is 2 / 11.</span>
5 0
2 years ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
2 years ago
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