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beks73 [17]
2 years ago
5

PLEASE HELP ASAP Prove that x+a is a factor of (x+a)^5 + (x+c)^5 + (a-c)^5

Mathematics
1 answer:
Brums [2.3K]2 years ago
6 0

P(x)=(x+a)^5 + (x+c)^5 + (a-c)^5

If x+a is a factor of P(x), then -a is a root of P(x).

Therefore

(-a+a)^5+(-a+c)^5+(a-c)^5=0\\\\0^5+(-1(a-c))^5+(a-c)^5=0\\\\(-1)^5(a-c)^5+(a-c)^5=0\\\\-(a-c)^5+(a-c)^5=0\\\\0=0

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Step-by-step explanation:

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James has a desk job and would like to become more fit, so he purchases a tread walker and a standing desk which will allow him
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answer is

-0.245 \pm2.160(0.205)

Step-by-step explanation:

After working this way for 6 months he takes a simple random sample of 15 days. He records how long he walked that day (in hours) as recorded by his fitness watch as well as his billable hours for that day as recorded by a work app on his computer.

Slope is -0.245

Sample size  n = 15

Standard error is 0.205

Confidence level 95

Sognificance level is (100 - 95)% = 0.05

Degree of freedom is n -2 = 15 -2 = 13

Critical Value =2.16 = [using excel = TINV (0.05, 13)]

Marginal Error = Critical Value * standard error

= 2.16 * 0.205

= 0.4428

-0.245 \pm2.160(0.205)

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A car manufacturer crash tests a certain model of car and measures the impact force. The test and model in question produce impa
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c.Approximately normal

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An arena receives $1,700 per event for 12 concerts. If the costs for this sponsorship total $14,000, what is the profit margin f
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The profit margin for the concert sponsorship is 31.3725 %

Step-by-step explanation:

An arena receives for 1 event = $1700

An arena receives for 12 events = 12 \times 1700

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The costs for this sponsorship total $14,000

Profit = 20400-14000

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Profit margin = 31.3725\%

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The marketing manager of a branch office of a local telephone operating company wants to study the characteristics of residentia
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Answer:

No. There is not enough evidence to support the claim that the population standard deviation is different from $12.

Step-by-step explanation:

The null hypothesis is that the true standard deviation is 12.

The alternative hypothesis is that the true standard deviation differs from 12.

We can state:

H_0: \sigma=12\\\\H_a: \sigma\neq12

The significance level is 0.10.

The sample size is n=15, so the degrees of freedom are:

df=n-1=15-1=14

The sample standard deviation is 9.25.

The test statistic is

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The critical values for rejecting the null hypothesis are:

\chi_{0.025,13}=5.00875\\\\\chi_{0.975,13}=24.7356

As T=8.32 is within the acceptance region (5.01, 24.74), the null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the population standard deviation is different from $12.

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