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2 years ago

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On the first day after the new moon, 2% of the Moon's surface is illuminated. On the second day, 6% is illuminated. Based on thi

s information, predict the day on which the Moon's surface is 50% illuminated and 100% illuminated.
illuminated.
PLEASE HELPPPPPPP!!!!!!!!!!!

2 answers:

maxonik [38]2 years ago

7 0

Answer:

50%: day 13

100%: day 26

Step-by-step explanation:

We are given two days and the amount of the moon that is illuminated. The two days are points on a straight line.

(1, 0.02), (2, 0.06)

y = mx + b

m = (0.06 - 0.02)/(2 - 1) = 0.04

y = 0.04x + b

0.02 = 0.04(1) + b

b = -0.02

y = 0.04x - 0.02

We want y = 50% = 0.5

0.04x - 0.02 = 0.5

0.04x = 0.52

x = 13

y = 100% = 1

0.04x - 0.02 = 1

0.04x = 1.02

x = 25.5

Rudiy272 years ago

7 0

Answer:

Day 13= 50%

Day 26=100%

Step-by-step explanation:

Answers vary. Sample response: A simple approach is to attempt a linear model starting at Day 1. If the illumination is increased by 4% every day, then after 11 more days (after Day 2) it reaches 50%. In 13 more days, illumination reaches 100%. This gives a prediction of Day 13 for 50% and Day 26 for 100%.

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The confidence interval would be given by:

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Step-by-step explanation:

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Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part a

Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

If we assume that we have $3$ groups and on each group from $j=1,\dots,6$ we have $6$ individuals on each group we can define the following formulas of variation:

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$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

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And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

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$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

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And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

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Part b

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The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

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