1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yan [13]
3 years ago
7

A violin student records the number of hours she spends practicing during each of nine consecutive weeks: 6.2 5.0 4.3 7.4 5.8 7.

2 8.4 1.2 6.3 Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:___________
Mathematics
1 answer:
spin [16.1K]3 years ago
5 0

Answer:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

Step-by-step explanation:

Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:

6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3

For this case we need to sort the data first on increasing way and we got:

1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4

For this case we have 9 values the median would be on the 5 position:

Median = 6.2

The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0

The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2

Then the interquartile rnage would be:

IQR = Q_3 -Q_1= 7.2-5=2.2

And 1.5 IQR = 1.5*2.2=3.3

The lower limit on this case would be:

LL= Q_1 -1.5 IQR= 5-3.3=1.7

And our value is 1.2 is lower than the lower limit 1.7

Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below  the first quartile.

False 1.2 is more than 1.5 IQR below the first quartile

C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  median.

False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median

D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  mean.

False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean

You might be interested in
I neeeee helppppppppp
kaheart [24]

Answer:

The pin number is 1147

Step-by-step explanation:

We have given that:

Maritza remembers her PIN because it is the product of two consecutive prime numbers that is between 1000 and 1500.

Pin number is between 1000 and 1500

PIN is the product of two consecutive prime numbers.

Prime numbers are those numbers which are divisible by one or itself only.

So,

prime numbers are 1, 3, 5, 7,………29, 31,37, 41, …..

Multiply  29 x 31 = 899  which does not come in the range of 1000 and 1500.

Next,

31 x 37 = 1147 this is between 1000 and 1500

Next

37 x 41 = 1517, which is above 1500.

Thus the pin number is 1147....

7 0
3 years ago
Raul calculated that he would spend $125 on school supplies this year. He actually spent $87.50 on school supplies. What is Raul
dybincka [34]

Answer:

C 30%

Step-by-step explanation:

87.50/125= 0.70

1-0.7-0.3

3 0
2 years ago
Read 2 more answers
15 - (-3) - 4<br><br> -16<br><br> 22<br><br> -8<br><br> 14
uranmaximum [27]
In this equation, you have to treat the number in the bracket first on the basis of BODMAS
15 - [-3]- 4
Note that when two minuses come together the product is a plus sign.
15 +3 - 4
You have to add before you subract
18 - 4 =14
Therefore, 15- [-3] - 4 = 14.
8 0
3 years ago
Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.
il63 [147K]

Answer:

The minimum value for  f(x) = 0.15(x + 1)^2 - 3 is -3.

Step-by-step explanation:

Given function is f(x) = 0.15(x + 1)^2 - 3

We need to find the maximum value or the minimum value for the function.

Now, differentiate f(x) = 0.15(x + 1)^2 - 3  w.r.t x.

f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\

f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)

Now, we will equate f'(x)=0 to find critical point.

0.3(x+1)=0\\x=-1

Plug this critical point in to the function f(x) = 0.15(x + 1)^2 - 3  we get,

f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3

Also, f''(x)=0.3 which is positive, We have minimum value.

So, the minimum value for  f(x) = 0.15(x + 1)^2 - 3 is -3.

7 0
3 years ago
Someone PLease please help me
makvit [3.9K]
Answer is 1 . Righttttttttttttt
4 0
3 years ago
Other questions:
  • Need help with double checking!
    7·1 answer
  • I will give 35points plus brainliest if right! please hurry!
    7·2 answers
  • Solve for x 3{x-2}+1=3x-5
    12·2 answers
  • What is the value of p?
    8·2 answers
  • Sean is a bookworm. On average, he reads pages in minutes. Sean's average unit rate for reading is pages per minute.
    14·2 answers
  • There are 5,280 feet in one mile. How many feet are there in 6 miles?
    5·1 answer
  • Need help pleaseeeeeeeeeeeeeer
    14·1 answer
  • Write an equation that represents the data in the table x 5,10,15,20 y 17,22,27,32
    9·1 answer
  • A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The s
    13·1 answer
  • Complete to get an equivalent fraction:<br> 9/c = _____/(3c)
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!