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Mumz [18]
3 years ago
10

Which scatterplot shows a linear association?

Mathematics
2 answers:
pickupchik [31]3 years ago
3 0

Answer:

The first one I believe

Step-by-step explanation:

Anna11 [10]3 years ago
3 0

Answer:

The answer is A

Step-by-step explanation:

The scatter plots likes b, c and d are non linear associations. You can easily remember this just by looking how weird shaping they are for example d. You can always remember that a scatter plot like d is always nonlinear.

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Maggie spent $50 buying 11 magazines.Some of the magazines cost $4 and the rest cost $5.How many $5 magazines did she buy?
antiseptic1488 [7]

Answer: She bought 6 $5 magazines and 5 $4 magazines

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we
kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

and \sigma_{{\overline}{Y}} = 0.38

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

X_{ij} = The number which comes up  on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

E(X_{ij}) = \frac {1 + 2 + 3 + 4 + 5 + 6}{6}

                            = 3.5       ∀ i = 1(1)10 and j = 1(1)20

and,

E(X^{2}_{ij} = \frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}

                                = \frac {1 + 4 + 9 + 16 + 25 + 36}{6}

                                = \frac {91}{6}

                                \simeq 15.166667

so, Var(X_{ij} = (E(X^{2}_{ij} - {(E(X_{ij})}^{2})

                                    \simeq 15.166667 - 3.5^{2}

                                    = 2.91667

   and \sigma_{X_{ij}} = \sqrt {2.91667}[/tex                                            [tex]\simeq 1.7078261036

Now we get that,

 Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}

We get that Y_{j}'s are iid RV's ∀ j = 1(1)20

Let, {\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}

      So, we get that E({\overline}{Y}) = E(Y_{j})

                                                                 = E(X_{ij}  for any i = 1(1)10

                                                                 = 3.5

and,

       \sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}}                                             = \frac {\sigma_{X_{ij}}}{\sqrt {20}}                                             = \frac {1.7078261036}{\sqrt {20}}                                            [tex]\simeq 0.38

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

and \sigma_{{\overline}{Y}} = 0.38

                                   

6 0
3 years ago
A robot can complete 8 tasks in 5/6 hour. Each task takes the same amount of time.
ludmilkaskok [199]

Answer:

A. 6.25

B. Nine tasks

Step-by-step explanation:

7 0
3 years ago
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Best hockey player hands down is?
stiv31 [10]

Answer:

Step-by-step explanation:

Sidney Crosby, Pittsburgh Penguins

6 0
2 years ago
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PLZZ HELP I NEED AN EXPERT ANSWER
IceJOKER [234]
This trend line shows a general pattern, and since we are with biological samples this trend line will not always be true.  The outliers could be from either a genetic difference or an environmental difference that occured in the persons life.
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3 years ago
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