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3 years ago

12

What do graphs help to see?

2 answers:

Greeley [361]3 years ago

8 0

Answer:

Time, distance, speed

bija089 [108]3 years ago

6 0

Answer:

Line graphs can also be used to compare changes over the same period of time for more than one group. Pie charts are best to use when you are trying to compare parts of a whole. They do not show changes over time. Bar graphs are used to compare things between different groups or to track changes over time.

Step-by-step explanation:

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3 years ago

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Which equation is the inverse of 5y+4= (x+3)² + 2?

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1 year ago

Counting numbers are to be formed using only the digits 1, 9, 8, 6, 2, 4, and 5. determine the number of different possibilities

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2 years ago

The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

2 years ago

A 20 foot ladder is leaning up against the side of a house the distance between the house and the base of the ladder is 4 feet f

Pie

Answer:

78.5°

Step-by-step explanation:

We solve for the above question, using the formula for the Trigonometric function of Cosine

cos θ = Adjacent/Hypotenuse

Adjacent = The distance between the house and the base of the ladder = 4 feet

Hypotenuse = Length of the Ladder = 20 feet

Hence,

cos θ = 4/20

θ = arc cos(4/20)

θ = 78.463040967°

Approximately = 78.5°

Therefore, the angle that the ladder makes with the ground is 78.5°

6 0

2 years ago