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Marianna [84]
3 years ago
10

What type of chart/graph must have all of its smaller parts add up to one whole?

Mathematics
1 answer:
adell [148]3 years ago
4 0
Pie chart Because it’s talking about one whole
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Answer a, b and c. See image below
dlinn [17]

Answer:

a) 3/5 < 4/5

b) In general if two fractions have the same denominator, then whichever fraction has the numerator closer to its denominator will be the largest fraction.

c)  \frac{7}{10} > \frac{9}{15}  <em>or</em>  \frac{7}{10}

Step-by-step explanation:

a) 3/5 < 4/5

Flip the sign and the placement of the fraction so 3/5 is less then 4/5.

b) In general if two fractions have the same denominator, then whichever fraction has the numerator closer to its denominator will be the largest fraction.

c) We need to change the denominators to a common denominator to compare the size of the two fractions:

\frac{7}{10} × \frac{3}{3} = \frac{21}{30}

\frac{9}{15} ×  \frac{2}{2} = \frac{18}{30}

The common denominators of the two fractions is 30. Comparing the two fractions:

\frac{21}{30} >\frac{18}{30}  <em>or</em>  \frac{18}{30}

so we get:  \frac{7}{10} > \frac{9}{15}  <em>or</em>  \frac{7}{10}

7 0
2 years ago
Show all work to identify the asymptotes and zero of the faction f(x) = 4x/x^2 - 16.
Reptile [31]

Answer:

  • asymptotes: x = -4, x = 4
  • zeros: x = 0

Step-by-step explanation:

The vertical asymptotes of the rational expression are the places where the denominator is zero:

  x^2 -16 = 0

  (x -4)(x +4) = 0 . . . . . true for x=4, x=-4

  x = 4, x = -4 are the equations of the vertical asymptotes

__

The zeros of a rational expression are the places where the numerator is zero:

  4x = 0

  x = 0 . . . . . . divide by 4

5 0
3 years ago
What is the area of the semicircle (use 3.14 for ρ)?
oee [108]

Answer:

127.17cm²

Step-by-step explanation:

Area of a semicircle: πr² ÷ 2

d = 18

r = 18/2 = 9

Area of a semicircle: 9²π ÷ 2 = 127.17cm²

3 0
3 years ago
Read 2 more answers
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.
spin [16.1K]

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

5 0
3 years ago
(45.22 – 73.91) – |65.85 – 79.15|
DerKrebs [107]

B. -41.99 is your answer.


Hope this helps & good luck. :)

8 0
3 years ago
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