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3 years ago

What display can be used to find how many pumpkins had masses below 6.0 kilograms. Histogram or box plot

Mathematics

1 answer:

Tamiku [17]3 years ago

7 0

Answer:

The correct option is Histogram.

Step-by-step explanation:

In statistics, we usually come across a vast set of data. Data can be explained well when it is presented as a graph rather than as a table since the graphs have the capability to make known a trend or comparison.

Different types of graphs used in statistics to describe data are:

Bar charts
Pie charts
Line charts
Histogram
Box Plot
Dot Plot

And so on.

The methods commonly used for depicting a frequency distribution are:

Histogram/Column graph
Bar graph
Frequency Polygon
Pie chart

A histogram is used to graphically represent the distribution of a random variable according to the frequency distribution.

Each bar of the histogram is constructed according to a range of values and represents the frequency for that particular range.

In this case, we need to find how many pumpkins had masses below 6.0 kilograms.

The histogram can be used to construct the graphically representation of the masses of pumpkins and their frequency.

The range can be decided according to the provided masses.

Thus, the correct option is Histogram.

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Question:

aivan3 [116]

Answer:

part A) The scale factor of the sides (small to large) is 1/2

part B) Te ratio of the areas (small to large) is 1/4

part C) see the explanation

Step-by-step explanation:

Part A) Determine the scale factor of the sides (small to large).

we know that

The dilation is a non rigid transformation that produce similar figures

If two figures are similar, then the ratio of its corresponding sides is proportional

Let

z ----> the scale factor

$\frac{CB}{C'B'}=\frac{CD}{C'D'}=\frac{BD}{B'D'}$

The scale factor is equal to

$z=\frac{CB}{C'B'}$

substitute

$z=\frac{4}{8}$

simplify

$z=\frac{1}{2}$

Part B) What is the ratio of the areas (small to large)?

Area of the small triangle

$A=\frac{1}{2}(2)(4)=4\ units^2$

Area of the large triangle

$A=\frac{1}{2}(4)(8)=16\ units^2$

ratio of the areas (small to large)

$ratio=\frac{4}{16}=\frac{1}{4}$

Part C) Write a generalization about the ratio of the sides and the ratio of the areas of similar figures

In similar figures the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In similar figures the ratio of its areas is equal to the scale factor squared

4 0

3 years ago

PLEASE HELP FAST!!!!!!!!!

tresset_1 [31]

B is definitely the answer

4 0

3 years ago

40 is what percent of 32?

Daniel [21]

Answer:

125%

Step-by-step explanation:

Is means equals and of means multiply

40 = P * 32 where P is in decimal form

Divide each side by 32

40/32 = P

1.25 = P

Now change to percent form ( multiply by 100)

125%

7 0

3 years ago

Read 2 more answers

Find all the real zeros of f(x)=3x^3-9x^2+3x-9 Can someone help me please

vaieri [72.5K]

Answer:

x=0,3

Step-by-step explanation:

start by dividing every term by 3

x^3-3x^2+x-3

group into 2 terms

(x^3-3x^2)+(x-3)

simplify as much as you can

x^2(x-3)+(x-3)

combine terms

(x^2)(x-3)

x=3, 0

4 0

3 years ago

In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

3 years ago