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Tju [1.3M]
3 years ago
10

Consecutive numbers follow one right after the other. An example of three consecutive numbers is 17, 18, and 19. Another example

is -100, -99, -98. How many sets of two or more consecutive positive integers can be added to obtain a sum of 100? What are they?
help pls
Mathematics
1 answer:
slamgirl [31]3 years ago
5 0
I would say none because if you try to find numbers that add up to 100 you won’t find them as integers. The answer is either no sets at all or you can do numbers that are consecutive like 32 1/3 , 33 1/3 and 34 1/3 . They all add up to 100 but I’m not sure cause they’re not full numbers so they’re not integers.
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AVprozaik [17]

Step-by-step explanation:

Angle formed at the circumference of the circle is half of the angle formed at the center of the same circle.

Hence,

m\angle ADC =  \frac{1}{2} \times  m\angle ABC \\  \\  \therefore \: m\angle ADC =  \frac{1}{2} \times  40 \degree \\  \\ \therefore \:m\angle ADC =  20 \degree  \\  \\

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A ′ B ′ C ′ triangle, A, prime, B, prime, C, prime is the image of △ A B C △ABCtriangle, A, B, C under a rotation about the orig
erma4kov [3.2K]

The angle of rotation of triangle ABC to A'B'C is 105°

<h3>How to illustrate the information?</h3>

It should be noted that the rotation from ABC to A'B'C is in a clockwise direction.

Point B is also on the x axis and point B' is in the second quadrant.

In this case, the angle that depicts the second quadrant is 105°.

In conclusion, the correct option is D.

The complete question is:

Triangle A’ B’ C’ is the image of A B C under a rotation about the origin, (0,0)

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B. -75 degrees

C. 75 degrees

D. 105 degrees

Learn more about triangles on:

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A new amusement park presold discounted tickets for the opening day as well as upon arrival at the park the opening day ticket s
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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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The value of the expression is 8. I hope this helps!!!

5 0
3 years ago
How does the speed of a runner vary over the course of a marathon (a distance of 42.195 km)? Consider determining both the time
Tom [10]

Answer:

The observed typical difference value of mode is 100 seconds approximately.

The proportion of runners ran the late distance more quickly than the early distance is approximately 1%

Step-by-step explanation:

Fundamentals

Consider that there are x favorable cases to an event E, out of a total of n cases. Then, the probability of that event is written as:

P(E)=  

Total number of cases /Number of favorable cases =  x/n

A histogram is constructed for continuous data, which is divided into classes called bins. The shape of the distribution can be determined from the histogram.

step 1

The provided histogram indicates time difference on xx -axis and frequency of runners on yy -axis. To determine the typical difference value, identify the peaks of the graph.

The histogram is skewed towards right side. The graph indicates that there are few outliers around 700 seconds. For a typical difference value, the value of mode is considered.

The graph indicates that the value of mode is 100 seconds approximately.

The observed typical difference value of mode is 100 seconds approximately.

Explanation

From the histogram, the typical difference value is obtained on the basis of guessing the value of mode which has been approximated to be around 100 seconds.

Step 2

From the histogram, it can be estimated that there are around 10 runners which has negative difference which means approxi8matley 10 runners ran the late distance more quickly than the early distance,

The approximate sample size can be calculated as:

Sample size=90+190+180+160+120+80+60+40+30+20

Thus, the proportion of runners is obtained as:

\begin{array}{c}\\p = \frac{{\left( \begin{array}{l}\\{\rm{Number of runners who has }}\\\\{\rm{negative time difference}}\\\end{array} \right)}}{{{\rm{Total sample size}}}}\\\\ = \frac{{10}}{{970}}\\\\ = 0.01\\\end{array}  

p=  <u> </u>Number of runners who has

<u>     negative time difference   </u>

      Total sample size

​  

=10/970

=0.001

The proportion of runners ran the late distance more quickly than the early distance is approximately 1%

EXPLANATION

The obtained proportion is 0.01. It indicates that there are approximately 1% of the runners who ran late distance more quickly than the early distance is very few.

​  

​  

3 0
3 years ago
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