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expeople1 [14]
1 year ago
12

A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade (x) and test grade (y)

are given in the accompanying table. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. Using this equation, find the projected test grade, to the nearest integer, for a student with a homework grade of 69.

Mathematics
1 answer:
Fiesta28 [93]1 year ago
6 0

Answer:

Il existe de nombreux exemples de portfolios d’artistes en ligne. Trouvez-en un et partagez un lien vers celui-ci. En vous basant sur le portfolio, écrivez quelques phrases évaluant l'artiste. S'il y a les réponses d'autres élèves sur le forum de discussion, examinez-les et voyez si vous êtes d'accord ou pas d'accord avec leur opinion sur le travail des artistes.

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The following is an incomplete paragraph proving that the opposite sides of parallelogram ABCD are congruent:
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<span>D) Angles BAC and DCA are congruent by the Alternate Interior Angles Theorem.</span>
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A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.
Dennis_Churaev [7]

Answer:

Margin of error at 90% is 0.024

Margin of error at 99% is 0.037

Step-by-step explanation:

Sample size = 1076

A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.

So, No. of adults said they believe ghosts = \frac{64}{100} \times 1076=688.64\sim 688

So, x = 688

n = 1076

\widehat{p} = \frac{x}{n}

\widehat{p} = \frac{688}{1076}

\widehat{p} = 0.639

ME=z \times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}

z at 90% confidence is 1.64

ME=1.64 \times \sqrt{\frac{0.639(1-0.639)}{1076}}

ME=0.024

So, margin of error at 90% is 0.024

Find the margin of error needed to be 99% confident.

z at 99% confidence is 2.58

ME=2.58 \times \sqrt{\frac{0.639(1-0.639)}{1076}}

ME=0.024

So, margin of error at 99% is 0.037

5 0
2 years ago
Ana played 555 rounds of golf, and her lowest score was an 808080.
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Answer:

c

Step-by-step explanation:

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2 years ago
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After a number of complaints about its tech assistance, a computer manufacturer examined samples of calls to determine the frequ
icang [17]

Answer:

a) Upper Control Limit = 0.10

   Lower Control Limit = 0.01

b) The tech assistance process is stable and in control.

Step-by-step explanation:

Given - After a number of complaints about its tech assistance, a

            computer manufacturer examined samples of calls to determine

            the frequency of wrong advice given to callers. Each sample

            consisted of 100 calls.

SAMPLE             1   2   3   4   5   6   7   8   9  10 11 12 13 14 15 16

Number of errors 5  3   5   7   4   6   8   4   5    9   3   4   5   6   6   7

To find - a. Determine 95 percent limits.

              b. Is the tech assistance process stable (i.e., in control)

Proof -

z = 95% confidence interval

  = 1.96

⇒z = 1.96

Proportion of defects,P = total defectives/total observations

                                     = 0.0544

⇒P = 0.0544

Now,

Q = 1-P

   = 0.9456

⇒Q = 0.9456

Now,

Average sample size, N = 100

Standard deviation = \sqrt{\frac{P.Q}{N} }

                             = 0.0227

Now,

Upper Control Limit = P + z(Standard deviation)

        = 0.0988

⇒Upper Control Limit = 0.0988 ≈ 0.10

And

Lower Control Limit = P - z(Standard deviation )

       = 0.0099

⇒Lower Control Limit = 0.0099 ≈ 0.01

∴ we get

Upper Control Limit = 0.10

Lower Control Limit = 0.01

b.)

Now,

it is clear that the fraction defective values are wit in upper an lower control limits.

So, The tech assistance process is stable and in control.

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