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tigry1 [53]
4 years ago
8

The sum of four consecutive odd integers is -72. Write an equation to model this situation, and ind the values of the four integ

ers.
Mathematics
1 answer:
Nutka1998 [239]4 years ago
3 0

Answer:

  • equation: (x-3) +(x-1) +(x+1) +(x+3) = -72
  • values: -21, -19, -17, -15

Step-by-step explanation:

When dealing with consecutive integers, it often simplifies the problem to work with their average value. We know the average value of the integers in this problem is the even integer between the middle two. We can call it x, and write the equation ...

  (x-3) +(x-1) +(x+1) +(x+3) = -72

This simplifies to ...

  4x = -72 . . . . . . . we knew this before we wrote the above equation, since the sum of 4 numbers is 4 times their average.

  x = -18 . . . . . . . . the middle number of the sequence

So, the numbers are:

  • -18-3 = -21
  • -18-1 = -19
  • -18+1 = -17
  • -18+3 = -15

_____

A more conventional approach is to define the variable as the integer at one end or the other of the sequence. If we make it be the lowest number, then the equation is ...

  (x) +(x +2) +(x +4) +(x +6) = -72

and that simplifies to ...

  4x +12 = -72 . . . . . collect terms

  4x = -84 . . . . . . . . subtract 12

  x = -21 . . . . . . . . . . divide by 4

Now, the other three numbers are found by adding 2, 4, and 6 to this one.

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each term of a sequence is 2 more than the previous term. if the third term is 8 find the first five terms of the sequence
soldier1979 [14.2K]

Step-by-step explanation:

T3=8

8=a+2d

where a is first term ,d is common ratio

8=a+2(2)

8=a+4

a=8-4

a=4

first term T1 is 4

T2=a+d

T2=4+2

T2=6

T4=a+3d

T4=4+3(2)

T4=4+6

T4=10

T5=a+4d

T5=2+4(2)

T5=4+8

T5=12

4,6,8,10,12

6 0
3 years ago
Please feel free to help if not i can figure it out, thank you.
Vika [28.1K]

Answer:

A makes the most sence cbwt

Step-by-step explanation:

8 0
3 years ago
A sample has a sample proportion of 0.3. which sample size will produce the
jasenka [17]

The sample size of 36 will produce the widest 95% confidence interval when estimating the population parameter option (b) is correct.

<h3>What are population and sample?</h3>

It is described as a collection of data with the same entity that is linked to a problem. The sample is a subset of the population, yet it is still a part of it.

We have:

A sample has a sample proportion of 0.3.

Level of confidence = 95%

At the same confidence level, the larger the sample size, the narrower the confidence interval.

As we have a 95% confidence interval the sample size should be lower.

The sample size from the option = 36 (lower value)

Thus, the sample size of 36 will produce the widest 95% confidence interval when estimating the population parameter option (b) is correct.

Learn more about the population and sample here:

brainly.com/question/9295991

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7 0
2 years ago
Find the sum of the first 10<br> terms.<br> 40,37,34,31,...
MArishka [77]

Answer:

It seems each number is 3 less than the previous number.

We must sum these 10 numbers

40, 37, 34, 31, 28, 25, 22, 19, 16, 13

The sum would equal the average of the middle 2 numbers times 10.

(28 + 25) / 2 = 53 / 2 = 26.5

We multiply by 10 because there are 10 numbers and this equals

265

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
.Here is a 90% confidence interval estimate of theproportion of adults who say that thelaw goes easy on celebrities: 0.645
bulgar [2K]

Answer:

The best estimate of the proportion of adults who say that the law goes easy on celebrities is 0.7435.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

In this question:

CI is between 0.645 and 0.842. So the best estimate of the proportion is:

(0.645+0.842)/2 = 0.7435

The best estimate of the proportion of adults who say that the law goes easy on celebrities is 0.7435.

5 0
3 years ago
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