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Tom [10]
2 years ago
13

Complete the ratio table with the numbers:

Mathematics
1 answer:
sammy [17]2 years ago
5 0

Answer:

C

24, 5, 48, 4 because blue is 6 times green

Step-by-step explanation:

Hope this helps

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A prism with a base area of 8 cm² and a height of 6 cm is dilated by a factor of 54 . What is the volume of the dilated prism? E
Aliun [14]

Question:

A prism with a base area of 8 cm² and a height of 6 cm is dilated by a factor of 5/4 .

What is the volume of the dilated prism?

Enter your answer, as a decimal, in the box.

Answer:

The volume of the dilated prism is 93.75 \ {cm}^{3}

Explanation:

A prism with a base area of 8 cm² and a height of 6 cm

The volume of the prism can be determined by the formula, V=Bh

Volume of the prism is given by

V=Bh

V=(8)(6)

V=46\ cm^3

Thus, the volume of the prism is 46 \ {cm}^{3}

It is also given that the volume of the dilated prism is dilated by a factor of \frac{5}{4}

Hence, the new volume is given by

Volume = 48(\frac{5}{4} )^3

             =48(\frac{125}{64} )

             =48(1.953125)

             =93.75 \ {cm}^{3}

Thus, the volume of the dilated prism is 93.75 \ {cm}^{3}

3 0
2 years ago
50 POINTS !!<br><br><br> PLEASE HELP AND EXPLAIN !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.
allochka39001 [22]
I need the point my point is -
4 0
3 years ago
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Evaluate 3^2 + (5 − 2) ⋅ 4 − 6 over 3.<br> 19<br> 16<br> 46<br> 24
Ludmilka [50]
3^2 +(5-2)* 4-6/3
Parenthesis, Exponents, Multiplication/Division, Addition/Subtraction
3^2+3*4- 6/3
9+3*4- 6/3
9+12- 6/3
9+12-2
9+10
19
So your answer is 19.
8 0
2 years ago
Riley has collected surveys for her final thesis project. One one question, 46.1% of respondents said that conditions were impro
ira [324]

Answer:

The maximum value of the confidence interval for this set of survey results is 51.73%.

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound.

These bounds depend on the sample proportion and on the margin of error.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this question:

Sample proportion: 46.1%

Margin of error: 5.63%.

Maximum value is the upper bound:

46.1+5.63 = 51.73

The maximum value of the confidence interval for this set of survey results is 51.73%.

5 0
3 years ago
Sorting through unsolicited e-mail and spam affects the productivity of office workers. An InsightExpress survey monitored offic
nydimaria [60]

Answer:

1. Refer to the explanation for the frequency table.

2. (a) 60%

   (b) 25%

Step-by-step explanation:

Part 1. The question is asking us to group the data according to the classes given and fill in the frequency (F), relative frequency (RF), cumulative frequency (CF) and relative cumulative frequency (RCF).

To compute the frequency for each class count the number of data that lies in the class range and write it. For the first class (1-5) we can see that there are 12 numbers which lie in the range 1 to 5. Those numbers are: 2, 4, 4, 1, 2, 1, 5, 5, 5, 3, 4, 4. Similarly, for the second class, the frequency is 3 because 8, 8, 8 lie in this range from our existing data. For the third class 11-15 there are 2 numbers in the given data which lie in this range and those numbers are 12, 15. Similarly, the rest of the frequencies can be computed.

For the relative frequency, the frequency of each class is divided by the total frequency i.e. 20.  

For the class 1-5, RF = 12/20 = 0.6.  

For 6-10, RF = 3/20 = 0.15

For 11-15, RF = 2/20 = 0.1

For 16-20, RF = 1/20 = 0.05

For 21-25, RF=1/20 = 0.05

For 26-30, RF = 0/20 = 0.00

For 31-34, RF= 1/20 = 0.05.

To compute the cumulative frequency, add the existing frequency of each class with the previous frequencies.

For 1-5, CF = 0+12 = 12

For 6-10, CF = 12+3=15

For 11-15, CF = 12+3+2 = 17

For 16-20, CF=12+3+2+1 = 18

For 21-25, CF = 12+3+2+1+1=19

For 25-30, CF = 12+3+2+1+1+0=19

For 31-34, CF = 12+3+2+1+1+0+1=20

Now, to compute relative cumulative frequency, add the existing relative frequency of each class with the previous relative frequencies.  

For 1-5, RCF = 0+0.6

For 6-10, RCF = 0.6 + 0.15 = 0.75

For 11-15, RCF = 0.6+0.15+0.1 = 0.85

The rest of the relative cumulative frequencies can be computed in the same way.

Class(Minutes)           F       RF     CF     RCF

      1-5                       12     0.60    12     0.60  

      6-10                      3     0.15     15     0.75  

      11-15                      2     0.10     17     0.85  

      16-20                    1      0.05    18     0.90  

      21-25                    1      0.05    19     0.95  

      26-30                   0     0.00    19     0.95  

      31-34                     1     0.05    20       1  

       <u>Total                  20</u>  

Part 2. Now, we are asked to compute the Ogive graph which is also called as the cumulative frequency graph. The cumulative frequency needs to be plotted on the y-axis and the upper limit of each class needs to be plotted on the x-axis. The graph is attached.

(a) From the graph we can see that the number of workers who spend less than 5 minutes on unsolicited e-mail and spam are 12. So the answer for this part is 12 workers.

Percentage = 12/20 x 100 = 60%

(b) From the graph we can see that the number of workers who spend less than 10 minutes on spam e-mail are 15. The question is asking for the number of people who spend more than 10 minutes. For this we need to subtract 15 from the total number of workers.  

Number of workers spending more than 10 minutes = 20-15 = 5 workers.

Percentage = 5/20 x 100 = 25%

4 0
3 years ago
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