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lora16 [44]
3 years ago
5

Five students wrote a mathematics test. The average mark was 68. If the marks for four of the students were 75, 62, 84 and 53, t

he mark of the fifth student was
Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

66

Step-by-step explanation:

Use the formula for the mean of a set of numbers:

                  sum of the numbers

average = --------------------------------

                     count of numbers

Here we know that the average mark was 68, so we can write:

               

                           75 + 62 + 84 + 53 + x

average = 68 = ----------------------------------

                                           5

or                            274 + x

                  68 = -------------------

                                    5

Solving for x, we multiply both sides by 5:     340 = 274 + x.

Subtracting 274 from both sides effectively isolates x:

 340 = 274 + x

- 274    -274

------------------------

  66      =    x

The fifth student's mark was 66.

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A research firm tests the miles-per-gallon characteristics of three brands of gasoline. Because of different gasoline performanc
Anni [7]

Answer:

A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.

B.)<u><em>The advantage of attempting to remove the block effect is</em></u>

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

Step-by-step explanation:

<em><u /></em>

<em><u>Using Two-way ANOVA method</u></em>

Given problem

<em><u>Observation              I          II       III          Row total (xr)</u></em>

A                              18 21 20            59

B                            24 26 27             77

C                            30 29 34             93

D                            22 25 24            71

<u>E                            20 23 24           63                      </u>

Col total (xc)             114 124 129        367

∑x²=9233→(A)

∑x²c/r

=1/5(114²+124²+129²)

=1/5(12996+15376+16641)

=1/5(45013)

=9002.6→(B)

∑x²r/c

=1/3(59²+77²+93²+71²+67²)

=1/3(3481+5929+8649+5041+4489)

=1/3(27589)

=9196.3333→(C)

(∑x)²/n

=(367)²/15

=134689/15

=8979.2667→(D)

Sum of squares total

SST=∑x²-(∑x)²/n

=(A)-(D)

=9233-8979.2667

=253.7333

Sum of squares between rows

SSR=∑x²r/c-(∑x)²/n

=(C)-(D)

=9196.3333-8979.2667

=217.0667

Sum of squares between columns

SSC=∑x²c/r-(∑x)²/n

=(B)-(D)

=9002.6-8979.2667

=23.3333

Sum of squares Error (residual)

SSE=SST-SSR-SSC

=253.7333-217.0667-23.3333

=13.3333

<u>ANOVA table</u>

Source                 Sums         Degrees      Mean Squares

of Variation       of Squares   of freedom

<u>                               SS                 DF              MS       F p-value</u>

B/ w     SSR=217.0667              4 MSR=54.2667    32.56 0.0001

rows

B/w     SSC=23.3333         c-1=2 MSC=11.6667        7 0.01

columns

<u>Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667                  </u>

<u>Total SST=253.7333 rc-1=14                                                        </u>

Conclusion:

<u> 1. F for between Rows</u>

The critical region for F(4,8) at 0.05 level of significance=3.8379

The calculated F for Rows=32.56>3.8379

Therefore H0 is rejected

<u>2. F for between Columns</u>

The critical region for F(2,8) at 0.05 level of significance=4.459

We see that the calculated F for Colums=7>4.459

therefore H0 is rejected,and concluded that there is significant differentiating between columns

<u><em>Part B:</em></u>

To analyze the data for completely  randomized designs click on anova two factor without replication  in the data analysis dialog box of the excel spreadsheet.

The following table is obtained

Source DF             Sum                  Mean           F Statistic

<u>                 (df1,df2)    of Square (SS) Square (MS)                    P-value</u>

Factor A       1 1496.5444 1496.5444 769.6514          0.001297

Rows

Factor B -     2 19.4444           9.7222               5                  0.1667

Columns

Interaction

AB               2    3.8889   1.9444        0.1013         0.9045

<u> Error     12   230.4            19.2                                           </u>

<u>Total 17 1750.2778 102.9575                                                         </u>

<u />

<u>Factor - A- Rows</u>

Since p-value < α, H0 is rejected.

<u>Factor - B- Columns</u>

Since p-value > α, H0 can not be rejected.

The averages of all groups assume to be equal.

<u>Interaction AB</u>

Since p-value > α, H0 can not be rejected.

<u><em>The advantage of attempting to remove the block effect is</em></u>

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

3 0
2 years ago
Math help please, thank you!
CaHeK987 [17]
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3 0
3 years ago
You are dealt one card from a 52-card deck. Find the probability that you are not dealt a non-picture card.
andreyandreev [35.5K]

Picture cards are Jacks, queens and Kings. There are 4 of each for a total of 12 picture cards.

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Probability of being dealt a non picture card would be 40/52

The probability of not being dealt a non picture card would be 1 - 40/52 = 12/52 = 3/13

Answer: 3/13

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g100num [7]

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Step-by-step explanation:

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