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Nuetrik
2 months ago
11

Select the correct answer. Twenty students in Class A and 20 students in Class B were asked how many hours they took to prepare

for an exam. The data sets represent their answers. Class A: {2, 5, 7, 6, 4, 3, 8, 7, 4, 5, 7, 6, 3, 5, 4, 2, 4, 6, 3, 5} Class B: {3, 7, 6, 4, 3, 2, 4, 5, 6, 7, 2, 2, 2, 3, 4, 5, 2, 2, 5, 6} Which statement is true for the data sets?
A. The mean study time of students in Class A is less than students in Class B.
B. The mean study time of students in Class B is less than students in Class A.
C. The median study time of students in Class B is greater than students in Class A.
D. The range of study time of students in Class A is less than students in Class B. E. The mean and median study time of students in Class A and Class B is equal.
Mathematics
1 answer:
zzz [12.3K]2 months ago
8 0

Response:

B. Students in Class B have a lower mean study time compared to those in Class A.

Detailed explanation:

To begin, we will calculate both the mean and median for each class.

Class A: 2, 5, 7, 6, 4, 3, 8, 7, 4, 5, 7, 6, 3, 5, 4, 2, 4, 6, 3, 5

sorted data: 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8

sum: 96

count: 20

Mean: 96 / 20 = 4.8

Median: In the sorted list, the 10th value is the first 5, so median = 5

Class B: 3, 7, 6, 4, 3, 2, 4, 5, 6, 7, 2, 2, 2, 3, 4, 5, 2, 2, 5, 6

Sorted data list: 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7

Sum: 80

Count: 20

Mean: 80 / 20 = 4

Median: In the sorted list, the 10th value is the first 4, so median = 4

Consequently, it is evident that Class B's average study time is less than that of Class A. Answer B.

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Each trapezoid in the figure below is congruent to trapezoid ABDC.
Inessa [12570]

Response:

B: 32 cm

To clarify, one should only sum the external measurements

6+6+4+4+3+3+3+3=32 cm

3 0
1 month ago
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The newly elected Chancellor of the Galactic Federation is interested in the opinions of all citizens regarding economic conditi
lawyer [12517]
Step-by-step approach: The Chancellor should implement a multi-stage sampling method for the survey. Initially, random samples of solar systems should be selected from the galaxy, followed by sampling planets within the chosen solar system, and finally, a random selection of countries within those planets.
7 0
1 month ago
1-i need to mix 25% mineral spirits to the varnish i am using what ratio of spirit to varnish am i using? 2-if i use 240 ml of v
Svet_ta [12734]

Answer:

(1) The necessary proportion of spirit to varnish stands at 1:3.

(2) A total of 80 ml of mineral spirit is required.

(3) The ratio of the heights to the widths of the tapestries is 3:2.

Step-by-step explanation:

(1) To achieve a mixture containing 25% mineral spirit, the blend consists of 25% spirit and 75% varnish.

Therefore, \frac{spirit}{varnish}=\frac{25}{75}

=\frac{1}{3}

Consequently, the proportion of spirit to varnish is 1:3.

(2) Using 240ml of varnish indicates this is 75% of the entire solution, which means the ratio of varnish to the total solution is 3:4.

Let the total solution quantity be x.

Thus, 240:x=3:4.

⇒\frac{240}{x}=\frac{3}{4}

⇒3x=240*4

⇒x=\frac{240*4}{3}

⇒x=320

This means the total solution amounts to 320 ml.

Now, calculating Spirit = Total solution - Varnish

⇒Spirit = 320ml-240ml

⇒Spirit = 80ml

Therefore, when using 240 ml of varnish, you will require 80 ml of mineral spirit.

(3) The dimensions of Robison's tapestries are uniform at 1.5m in length and 1m in width. Initially, to obtain whole numbers, we multiply these dimensions by 10.

Thus, the dimensions become 15m long and 10m wide.

Now, the proportion of Height to Width is 15:10

⇒\frac{Height}{Width} =\frac{15}{10}

⇒\frac{Height}{Width} =\frac{3}{2}

Thus, the proportion of the heights of the tapestries to their widths is 3:2.

8 0
1 month ago
A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)
AnnZ [12381]

Response:

a. As student debt rises, current investment diminishes.

b. Y= 68778.2406 - 1.9112X

For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.

c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step breakdown:

Hello!

Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:

Y: Current investment by an individual who graduated from college five years prior.

X: Total debt of an individual upon graduating five years ago.

a)

To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.

The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.

Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.

b)

The population regression equation is Y= α + βX +Ei

To develop this equation, estimates for alpha and beta are required:

a= Y[bar] -bX[bar]

a= 44248.55 - (-1.91)*12829.70

a= 68778.2406

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }

b=\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }

b= -1.9112

∑X= 256594

∑X²= 4515520748

∑Y= 884971

∑Y²= 43710429303

∑XY= 9014653088

n= 20

Averages:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation becomes:

Y= 68778.2406 - 1.9112X

For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.

c)

To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)

Using t=  b - β  =  -1.91 - 0  = -31.83

                 Sb         0.06

The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001

Since this P-value is underneath the significance level, we reject the null hypothesis.

In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.

The P-value remains: 0.0001

Using this method, we similarly reject the null hypothesis.F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37

In conclusion, at a significance level of 1%, there exists a linear relationship linking current investment to college debt.

The accurate statement is:

There exists a significant linear association between college debt and current investment since the P-value is less than 0.1.

d)

To forecast the value of Y when X is set, it is essential to substitute X in the estimated regression equation.

Y/$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The anticipated investment for someone with a college debt of $5000 is $59222.2406.

e)

To determine the proportion of variation in the dependent variable that the independent variable accounts for, the coefficient of determination R² must be calculated.

R²= 0.9818

R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }

R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }

This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X

I trust this is beneficial!

5 0
1 month ago
In regional spelling bee, the 8 finalists consist of 3 boys and 5 girls. Find the number of sample point the sample space S for
PIT_PIT [12445]

Answer:

i) A total of 40320 different arrangements

ii) For the initial 3 spots, there are 336 different combinations.

Step-by-step explanation:

Given: The total finalists = 8

The count of boys = 3

The count of girls = 5

To determine the number of sample point in the sample space S for possible arrangements, we calculate the factorial of 8!

The number of possible arrangements equals 8!

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8

= 40320

ii) Among the 8 finalists, we must select the first 3 spots. The sequence matters, hence we utilize permutation.

nPr =\frac{n!}{(n - r)!}

Here n = 8 and r = 3

Substituting n = 8 and r = 3 into the formula, we arrive at

8P3 = \frac{8!}{(8 - 3)!}

= \frac{8!}{5!} \\= \frac{1.2.3.4.5.6.7.8}{1.2.3.4.5}

= 6.7.8

= 336

Thus, there are 336 different arrangements for the first 3 spots.

3 0
2 months ago
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