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Allushta [10]
3 years ago
10

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

i) for the first graph the relation of the scatter-plot is non-linear

ii) for the second graph the relation of the scatter-plot is linear and negative

iii) for the third graph the relation of the scatter-plot is linear and positive.

iv) for the fourth graph the relation of the scatter plot is non-linear

Step-by-step explanation:

i) for the first graph the relation of the scatter-plot is non-linear

ii) for the second graph the relation of the scatter-plot is linear and negative

iii) for the third graph the relation of the scatter-plot is linear and positive.

iv) for the fourth graph the relation of the scatter plot is non-linear

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2 years ago
A hockey team is convinced that the coin used to determine the order of play is weighted. The team captain steals this special c
fredd [130]

Answer:

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

Step-by-step explanation:

Let p be the probability of heads in a single toss of the coin. Then our null hypothesis that the coin is fair will be formulated as

H0 :p 0.5   against   Ha: p ≠ 0.5

The significance level is approximately 0.05

The test statistic to be used is number of heads x.

Critical Region: First we compute the probabilities associated with X the number of heads using the binomial distribution

Heads (x)        Probability (X=x)                        Cumulative     Decumulative

0                        1/16384 (1)             0.000061     0.000061

1                         1/16384  (14)         0.00085             0.000911

2                       1/16384 (91)           0.00555             0.006461

3                       1/16384(364)         0.02222

4                       1/16384(1001)         0.0611

5                       1/16384(2002)       0.122188

6                        1/16384(3003)      0.1833

7                         1/16384(3432)      0.2095

8                        1/16384(3003)       0.1833

9                        1/16384(2002)       0.122188

10                       1/16384(1001)        0.0611

11                       1/16384(364)        0.02222

12                      1/16384(91)            0.00555                             0.006461

13                     1/16384(14)              0.00085                           0.000911

14                       1/16384(1)            0.000061                            0.000061

We use the cumulative and decumulative column as the critical region is composed of two portions of area ( probability) one in each tail of the distribution. If  alpha = 0.05 then alpha by 2 - 0.025 ( area in each tail).

We observe that P (X≤2) =   0.006461 > 0.025

and

P ( X≥12 ) = 0.006461 > 0.025

Therefore true significance level is

∝=  P (X≤0)+P ( X≥14 ) = 0.000061+0.000061= 0.000122

Hence critical region is (X≤0) and ( X≥14)

Computation x= 12

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

3 0
3 years ago
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