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vovangra [49]
3 years ago
11

find the sum of all number from 50 to 350 which are divisible by 6. hence, find the 15th term of that A.P​

Mathematics
1 answer:
QveST [7]3 years ago
8 0

Step-by-step explanation:

this is the answer shreya sorry but attachment mein second se start hai answer

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Is 11,13,17 a right triangle
11Alexandr11 [23.1K]

Answer:

no

Step-by-step explanation:

We can use the Pythagorean theorem to check

a^2 +b^2 = c^2

11^2+13^2 = 17^2

121+169 =289

410 does not equal 289

This is not a right triangle

3 0
3 years ago
Read 2 more answers
Describe a real-life multiplication situation for which an estimate makes sense . Explain why it makes sense
monitta

Answer: Here we have a real life multiplication situation for which an estimate make sense:-


Miley has 212 fiction books. Sharon has 3 times fiction books has many as Miley has, how many fiction books Sharon has

Solution :- Number of fiction books Miley has = 212

Number of fiction books Sharon has= 3 times 212

We know that 212 is near to 200 , so round off 212 to 200 (to the nearest hundred)

Thus the product becomes 3\times200=600

Therefore, the estimated number of fiction books Sharon has=600





7 0
3 years ago
5)
Natalija [7]

Answer:

i think the 2nd one

Step-by-step explanation:

7 0
2 years ago
Pick a city in Maryland and determine the average high temperatures for each
Elena-2011 [213]

<span>Baltimore Average Temperature 2016</span>

<span>
</span>

<span>Attached image.</span>

5 0
3 years ago
Read 2 more answers
Identify the class​ width, class​ midpoints, and class boundaries for the given frequency distribution.
Crank

Answer:

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

c.

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

M=\frac{lower class limit+upper class limit}{2}

Class

Interval Midpoints

45-48 \frac{45+48}{2}=46.5

49-52 \frac{49+52}{2}=50.5

53-56 \frac{53+56}{2}=54.5

57-60 \frac{57+60}{2}=58.5

61-64 \frac{61+64}{2}=62.5

65-68 \frac{65+68}{2}=66.5

69-72 \frac{69+72}{2}=70.5

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

7 0
3 years ago
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