All categories

2 years ago

The first four terms of an arithmetic sequence are 28, 20, 12, and 4. which term of the sequence is -36

1 answer:

7 0

For an arithmetic sequence, the nth term is the first term plus (n-1)×the differece
the first term is 28, the difference between each adjacent number is 8
so -36=(28)+8(n-1)
8(n-1)=64
n-1=8
n=9
the 9th term is -36

You might be interested in

1. A sample consists of 75 TV sets purchased several years ago. The replacement times of those TV sets have a mean of 8.2 years.

Greeley [361]

Answer:

8.2+/-0.25

= ( 7.95, 8.45) years

the 95% confidence interval (a,b) = (7.95, 8.45) years

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 8.2 years

Standard deviation r = 1.1 years

Number of samples n = 75

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

8.2+/-1.96(1.1/√75)

8.2+/-1.96(0.127017059221)

8.2+/-0.248953436074

8.2+/-0.25

= ( 7.95, 8.45)

Therefore the 95% confidence interval (a,b) = (7.95, 8.45) years

3 0

2 years ago

Don't get this question

solniwko [45]

Answer:

17 and 18

Step-by-step explanation:

Writing every 2 digit number gives us :

10-99 including 10 and 99

For the first one we are looking for numbers where the number 3 appears only once so 33 would be invalid.

13,23,30,31,32,34,35,36,37,38,39,43,53,63,73,83,93

There are 17 2-digit numbers that have 3 exactly once

For the second one we are looking for number where the number 3 appears a minimum of once so 33 would be valid :

13,23,30,31,32,33,34,35,36,37,38,39,43,53,63,73,83,93

There are 18 2-digit numbers that have 3 at least once

Hope this helped and have a good day

5 0

1 year ago

Seven and thirty twp hundreths plus two and one tenth times two and seven tenths minus 18a

alexdok [17]

Answer: Write the expression as:

" 70 $\frac{32}{100}$ +
2 $\frac{1}{10}$ * 2 $\frac{7}{10$ − 18a " ;
______________________________________________________
or; write as:
______________________________________________________
" 70.32 + (2.1) * (2.7) − 18a " ;
______________________________________________________

To simplify:
______________________________________________________

Using "PEDMAS" (the "order of operations") ;

the "multiplication" comes first;

So: → "(2.1) * (2.7) = 5.67 " .

And rewrite:
______________________________________________________
" 70.32 + 5.67 − 18a " .

Now: " 70.32 + 5.67 = 75.99 " ;

So, we can the final simplified expression as:
______________________________________________________
" 75.99 − 18a " ;

or; write as: " 75 $\frac{99}{100}$ − 18a " .
______________________________________________________

7 0

2 years ago

Please help math

Naddika [18.5K]

$\displaystyle\\
15(2a - 2) = 5(a^2 -1) \\ \\ 
30a - 30 = 5a^2-5\\\\
-5a^2 + 30a -30 + 5 =0\\\\
-5a^2 + 30a -25 =0 ~~~~~\Big| \times (-1) \\\\
5a^2 - 30a +25 = 0~~~~~\Big| : 5 \\\\
a^2 - 6a +5 = 0 \\ \\ 
a_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{6\pm \sqrt{36-20}}{2}= \frac{6\pm \sqrt{16}}{2}=\frac{6\pm 4}{2}= \boxed{3\pm 2} \\ \\ 
a_1 = 3+ 2 = \boxed{5}\\\\
a_2 = 3-2 = \boxed{1}$

8 0

2 years ago

Read 2 more answers

Find an equation for the line that passes through the point (-3, 6), parallel to the line through the points (0,-7) and (4,-15).

Vitek1552 [10]

Broken down into steps:

1. Find the slope of the line segment that connecs the points (0,-7) and (4,-15).

2. Start with the point-slope formula for the equation of a straight line:

y-b = m(x-a), where the given point is (a,b). Borrow the value of m that you calculated in (1), above, and insert it into this point-slope formula.
Finish up by subst. the x- and y-values in (-3,6) into this formula.

Done!

You could, of course, solve this result for y if you wished.

8 0

2 years ago