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professor190 [17]
2 years ago
7

What is the difference of the expression below? (12f-8g+3h) - (4f - g +5h)

Mathematics
1 answer:
Furkat [3]2 years ago
7 0

<u>Answer: </u>

The difference of (12f - 8g + 3h)  -(4f – g + 5h) is 8f - 7g - 2h

<u>Solution: </u>

From question given that the expression is

(12f - 8g + 3h) - (4f – g + 5h)

We have to find the difference of both the expressions.

By removing the parenthesis the above equation becomes,

12f - 8g + 3h - 4f + g - 5h

Separate the terms of f, g and h in above equation,

12f - 4f - 8g + g + 3h - 5h

On simplifying the above expression,

8f - 7g - 2h

Hence difference of (12f - 8g + 3h) - (4f – g + 5h) is 8f - 7g - 2h  

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The shuttle bus from your parking lot and your office building operates on a 15 minute schedule. You arrive at the parking lot a
mylen [45]

Answer:

(a) The standard deviation of your waiting time is 4.33 minutes.

(b) The probability that you will have to wait more than 2 standard deviations is 0.4227.

Step-by-step explanation:

Let <em>X</em> = the waiting time for the bus at the parking lot.

The random variable <em>X</em> is uniformly distributed with parameters <em>a</em> = 0 to <em>b</em> = 15.

The probability density function of <em>X</em> is given as follows:

f_{X}(x)=\frac{1}{b-a};\ a

(a)

The standard deviation of a Uniformly distributed random variable is given by:

SD=\sqrt{\frac{(b-a)^{2}}{12}}

Compute the standard deviation of the random variable <em>X</em> as follows:

SD=\sqrt{\frac{(b-a)^{2}}{12}}

      =\sqrt{\frac{(15-0)^{2}}{12}}

      =\sqrt{\frac{225}{12}}

      =\sqrt{18.75}\\=4.33

Thus, the standard deviation of your waiting time is 4.33 minutes.

(b)

The value representing 2 standard deviations is:

X=2\times SD=2\times4.33=8.66

Compute the value of P (X > 8.66) as follows:

P(X>8.66)=\int\limits^{10}_{8.66} {\frac{1}{15-0}}\, dx\\

                    =\frac{1}{15}\times \int\limits^{10}_{8.66} {1}\, dx\\

                    =\frac{1}{15}\times |x|^{15}_{8.66}\\

                    =\frac{15-8.66}{15}\\

                    =0.4227

Thus, the probability that you will have to wait more than 2 standard deviations is 0.4227.

4 0
1 year ago
ben earns $9 per hour and $6 for each delivery he makes. he wants to earn more than $155 in an 8-hour workday. what is the least
Evgen [1.6K]
The answer would be 13
9x8=72
155-72=83
83 / 6= 13
5 0
1 year ago
Read 2 more answers
The subjects of a study by Dugoff et al. (A-5) were 10 obstetrics and gynecology interns at the University of Colorado Health Sc
astra-53 [7]

Answer:

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

Step-by-step explanation:

Intern            No. of Breast

Number        Exams Performed               X²

1                         30                                  900

2                        40                                 1600

3                        8                                        64

4                        20                                   400

5                          26                                 676

6                           35                               1225

7                            35                               1225

8                            20                                400

9                             25                              625

<u>10                                 20                        400 </u>

<u>                                   </u><u> ∑ 259                  ∑ 7515</u>

Mean= X`= ∑x/n= 259/10= 25.9

Variance = s²= 1/n-1[∑X²- (∑x)²/n]

= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]

= 806.9/9=89.655= 89.66

Standard Deviation= √89.655= 9.4687

Hence

The value of t with significance level alpha= 0.05 and 9 degrees of freedom  is t(0.025,9)= 2.262

The 95 % Confidence interval is given by

x`±t(∝,n-1) s/√n

So Putting the values

25.9± 2.262( 9.4687/√10)

= 25.9 ±2.262 (2.9943)

= 25.9 ± 6.7730

= 25.9 +6.7730=32.6730

25.9 -6.7730= 19.1269

= 19.1269, 32.6730

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

6 0
2 years ago
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
2 years ago
Read 2 more answers
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Vikentia [17]
The answer is 8 sugar cubes
8 0
2 years ago
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