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melamori03 [73]
4 years ago
10

A right triangle has a hypotenuse of 70 feet and a leg of 35 feet. What is the length of the other leg? Select one:

Mathematics
1 answer:
saveliy_v [14]4 years ago
3 0
H = 70ft.
P = 35ft.
B = ?

H^2 = P^2 + B^2
70×70 = (35×35) + B^2
B^2 = 4900-1225
B = √3675
B = 35√3 ft.
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a teacher promised a movie day to the class that did better, on average, on their test. The box plot shows the results of the te
natka813 [3]

The question is incomplete. The complete question is

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.

Which class should get the reward, and why?

- The 2nd-period class should get the reward. They have the highest score, a perfect 100.

- The 2nd-period class should get the reward. They have a higher median.

- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.

The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is

\frac{77+78+89+92+100}{5}=87.2

In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.

Hence, the mean of the results of the 4th period is

\frac{72+83+89+96+98}{5}=87.6

Hence, obviously, the 4th period is having more mean.

Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.

So, just by observing the box plot, we can give the statement that  "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "

Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

Learn more about box plots here-

brainly.com/question/1523909

#SPJ10

6 0
2 years ago
The following data represent the pulse rates? (beats per? minute) of nine students enrolled in a statistics course. Treat the ni
charle [14.2K]

Answer:

(a)77.4bpm

(b)Mean of Sample 1 = 70.3 beats per minute.  

Mean pulse of sample 2 = 70 beats per minute.

(c)

  • The mean pulse rate of sample 1 underestimates the population mean.
  • The mean pulse rate of sample 2 underestimates the population mean.

Step-by-step explanation:

(a)Population mean pulse.

The pulse of the nine students which represent the population are:

  • Perpectual Bempah      64
  • Megan Brooks               77
  • Jeff Honeycutt               89
  • Clarice Jefferson           69
  • Crystal Kurtenbach       89
  • Janette Lantka              65
  • Kevin McCarthy             88
  • Tammy Ohm                  69
  • Kathy Wojdya                 87

\text{Population Mean} =\dfrac{64+77+89+69+89+65+88+69+87}{9} \\=\dfrac{697}{9} \\\\=77.44

The population mean pulse is approximately 77.4 beats per minute.

(b)Sample 1: {Janette,Clarice,Megan}

  • Janette: 65bpm
  • Clarice: 69bpm
  • Megan: 77bpm

Mean of Sample 1

\text{Sample 1 Mean} =\dfrac{65+69+77}{3} \\=\dfrac{211}{3} \\\\=70.3

Sample 2: {Janette,Clarice,Megan}

  • Perpetual: 64bpm
  • Clarice: 69bpm
  • Megan: 77bpm

Mean of Sample 2

\text{Sample 2 Mean} =\dfrac{64+69+77}{3} \\=\dfrac{210}{3} \\\\=70

The mean pulse of sample 1 is approximately 70.3 beats per minute.  

The mean pulse of sample 2 is approximately 70 beats per minute.

(c)

  • The mean pulse rate of sample 1 underestimates the population mean.
  • The mean pulse rate of sample 2 underestimates the population mean.
7 0
3 years ago
Please do the question. If you don't know don't answer!
GarryVolchara [31]

Answer:

5

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
If t=0 in 1980, find the value of A. Round answer to the nearest million.
balu736 [363]

P=Ae^kt

225 = 210 * e^(k*(1990-1980)

225/210=e^10k ln(225/210)=10k

k=ln(225/210)/10=0.0069

P = 210*e^(0.0069t)

for 2000 ===> t = 2000-1980=20

P = 210*e^(0.0069*20)

P=241.0749=241

4 0
4 years ago
Can someone help me plz
enyata [817]

Answer:the answer is 420 :)



Step-by-step explanation:


5 0
3 years ago
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