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Hunter-Best [27]
3 years ago
11

If a + 1 = b, then b > a.

Mathematics
2 answers:
photoshop1234 [79]3 years ago
8 0
<h2>Answer with explanation:</h2>

A)

When a is positive:

     Let us consider a=10

      and a+1=b

       This implies that:

          b=11

     As we know that 11>10

           Hence, b>a

When a is negative:

   Let us consider a= -2

      and a+1=b

       This implies that:

          b= -1   (Since -2+1= -1 )

     As we know that -1 > -2

           Hence, b>a

B)

No, it is not possible to create a set of values for a and b such that the numerical relationship shown in the given conditional statement is false .

Since as we no that no matter whether a is positive or negative we always have:

                        a < a+1

Hence,      a< b  ( Since b=a+1 )

for all the values of  a and b.

kogti [31]3 years ago
4 0
If a=-10;then:
a+1=-10+1=-9=b
-9>-10

if a=10; then
a+1=10+1=11=b
11>10
A)The pair of values for a and be are: a=-10, then the value of b would be: -9;
And a=10; then the value of b would be:11;

B) It isn´t possible to create a pair of values for a and be, in wich the numerical relationship shown in the given conditional stament is false, therefore b>a if a+1=b
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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

                                   

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