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4 years ago

Write the sum using summation notation assuming the suggested pattern continues. -8-3+2+7+...+67

1 answer:

5 0

You can use the generic representation of a term of an arithmetic sequence:
$a_n=a_1+d(n-1)$
Filling in the values a₁=-8, d=5, you have
$a_n=-8+5(n-1)\\a_n=-13+5n$

Then the value of n for the last term can be found as
67 = -13 + 5n
80 = 5n
16 = n
and the sum can be written as

$S_{16}=\sum\limits_{n=1}^{16}{(-13+5n)}$

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a chess player played 136 games total. If he won 56 of the games, what is the ratio of games he lost to games he won?

wolverine [178]

Answer:

80:56

Step-by-step explanation:

if you do 136-56, you get 80. therefore the number of games lost is 80. you already know he won 56, so the ratio is 80:56.

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3 years ago

A grocery store manager received two shipments of oranges. According to the distributor of the oranges, the oranges in both ship

Volgvan

The answer is
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7 0

3 years ago

Read 2 more answers

What is m in 1/3m = -4

ahrayia [7]

Answer:

m=-12

Step-by-step explanation:

Let's solve your equation step-by-step.

1 / 3 m= −4

Step 1: Multiply both sides by 3.

3*( 1 / 3 m) = (3) * (−4)

m=−12

3 0

3 years ago

A car initially traveling at 20 kilometers per hour takes 10 seconds to speed up to 70 km/hour. At what rate is it changing spee

Snezhnost [94]

Initial velocity=u=20km/h
Final velocity=v=70km/h
Time=10s=t

Convert speeds to m/s

$\\ \rm\longmapsto u=20\times\dfrac{5}{18}=5.05m/s$

$\\ \rm\longmapsto v=70\times \dfrac{5}{18}=19.4m/s$

$\boxed{\sf Acceleration=\dfrac{v-u}{t}}$

$\\ \rm\longmapsto Acceleration=\dfrac{19.4-5.05}{10}$

$\\ \rm\longmapsto Acceleration=\dfrac{14.3}{10}$

$\\ \rm\longmapsto Acceleration=143m/s^2$

7 0

3 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago