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ICE Princess25 [194]
3 years ago
13

The following data provides the number of calories of various types of frozen bars{25,35,200,280,80,80,90,40,45,50,50,60,90,100,

120,40,45,60,70,350}
a.Draw a histogram to represent the datab.Find the measures of center.

c.Can you find the measures of center only from the histogram? Explain.

Mathematics
1 answer:
Korolek [52]3 years ago
5 0
A] The histogram of the data set given is shown on the attachment.
(<span>25,35,200,280,80,80,90,40,45,50,50,60,90,100,120,40,45,60,70,350</span>)
Measures of the center are:
25,75,125,175,225, 275, 325
The histogram is J-shaped

b] The measures of centers can on be obtained through Histogram, this is because they are not random numbers, they have to be generated from the histogram.

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What is the ratio of a : b, knowing that a — b = 3 <br> a + b = 5
UNO [17]

Answer:

4:1

Step-by-step explanation:

a — b = 3

then a= 3+b

a + b = 5

3+b +b = 5

3+2b=5

2b=5-3

2b=2

b=1

a= 3+b

a=3+1=4

a:b= 4:1

3 0
3 years ago
Find the value of b.
Nesterboy [21]

Answer: 230^{\circ

Step-by-step explanation:

For convenience, I labeled some points as shown in the attached picture.

Also, I assume \overline{AB} and \overline{BC} are tangents to the circle.

  1. \overline{BO} \cong \overline{BO} (reflexive property)
  2. \overline{AB} \cong \overline{BC} (tangents drawn from a common external point are congruent)
  3. \angle BAO \cong \angle BCO (right angles are congruent)

Therefore, we know \triangle ABO \cong \triangle CBO by HL.

Thus, by CPCTC,

\angle AOB \cong \angle BOC \implies m\angle BOC=65^{\circ}\\\\\implies m\angle AOC=130^{\circ}

This means the measure of minor arc AC is 130^{\circ}, and thus b=360^{\circ}-130^{\circ}=\boxed{230^{\circ}}

7 0
1 year ago
A bathtub is draining at a constant rate. After 2 minutes, it holds 30 gallons of water. Three minutes later, it holds 15 gallon
Ivahew [28]

Answer:

60-15y=x

Step-by-step explanation:

3 0
3 years ago
Algebra help please I need a equation please
Rudik [331]
K = s + 4
k + s = 26
s + 4 = 26
Subtract 4 from both sides of the equation.
s = 22
Steve ran 22 miles.
3 0
3 years ago
A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream. What is the rate of
nikitadnepr [17]
<h3>Rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr</h3><h3><u>Solution:</u></h3>

Given that,

A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream

Therefore,

Upstream distance = 165 km

Upstream time = 3 hours

<h3><u>Find upstream speed:</u></h3>

speed = \frac{distance}{time}\\\\speed = \frac{165}{3}\\\\speed = 55

Thus upstream speed is 55 km per hour

Downstream distance = 510 km

Downstream time = 6 hours

<h3><u>Find downstream speed:</u></h3>

speed = \frac{510}{6}\\\\speed = 85

Thus downstream speed is 85 km per hour

<em><u>If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then</u></em>

Speed downstream = u + v km/hr

Speed upstream = u - v km/hr

Therefore,

u + v = 85 ----- eqn 1

u - v = 55 ----- eqn 2

Solve both

Add them

u + v + u - v = 85 + 55

2u = 140

u = 70

<em><u>Substitute u = 70 in eqn 1</u></em>

70 + v = 85

v = 85 - 70

v = 15

Thus rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr

3 0
3 years ago
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