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leva [86]
3 years ago
11

How many terms are in the arithmetic sequence 7, 1, −5, …, −161?

Mathematics
1 answer:
uranmaximum [27]3 years ago
5 0

Answer:

29

Step-by-step explanation:

The n th term of an arithmetic sequence is

• a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

d = 1 - 7 = - 5 - 1 = - 6 and a₁ = 7, a_{n} = - 161, hence

7 - 6(n - 1) = - 161 ← solve for n

7 - 6n + 6 = - 161

13 - 6n = - 161 ( subtract 13 from both sides )

- 6n = - 174 ( divide both sides by - 6 )

n = 29

There are 29 terms in the sequence

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HELP PLEASE !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
aleksandrvk [35]

50km to 80km is 30km

10c to -80c is -90c

and since it linear (Which means the change is constant or the same)

every 10 km show a -30c decrease

so

60 km is -20c

70km is -50c


answer is c

5 0
3 years ago
Can anyone help please?
stepladder [879]

Answer:

h(t) = -16t(t-6)

h(2) = 128

Step-by-step explanation:

h(t) = -16t² + 96t

h(t) = -16t(t-6)

t = 3

h(2) = -16(2)(2 - 6)

h(2) = 128

7 0
3 years ago
What is the slope of line LM given L(9, -2) and M(3, -5)?
jasenka [17]

Answer:

1/2

Step-by-step explanation:

(y2-y1)/(x2-x1) is the equation for finding slope with two given points

-5-(-2) over 3-9 equals

-3/-6=-1/-2=1/2

7 0
4 years ago
A certain arithmetic sequence has this explicit formula for the nth term:
Margaret [11]
The number 6 could replace the gap :)
6 0
3 years ago
Read 2 more answers
Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that th
11111nata11111 [884]

Answer:

The null hypothesis is H_0: \mu = 14

The alternate hypothesis is H_a: \mu < 14

The test statistic is t = -1.95.

The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the  highlighters wrote for less than 14 continuous hours.

Step-by-step explanation:

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.

At the null hypothesis, we test if the mean is 14 hours, that is:

H_0: \mu = 14

At the alternate hypothesis, we test if the mean is less than 14 hours, that is:

H_a: \mu < 14

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that \mu = 14

X = 13.6 hours, s = 1.3 hours. Sample of 40:

In addition to the values of X and s given, we have that n = 40

Test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}

t = -1.95

The test statistic is t = -1.95.

P-value:

The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.

Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the  highlighters wrote for less than 14 continuous hours.

5 0
3 years ago
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