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3 years ago

The radius of a circle is

Mathematics

2 answers:

Inessa05 [86]3 years ago

5 0

Red box
measurement from the center of circle to the edge of a circle

svetlana [45]3 years ago

5 0

Answer:

The measurement from the center of a circle to an edge of the circle

Step-by-step explanation:

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a recipe asks that the following three ingredients be mixed together as follows: add 1/2 of a cup of flour for every 1/2 of a te

Harman [31]

Answer:

Step-by-step explanation:

7 0

3 years ago

An airline charges $(22x + 20) for a ticket, $(x + 1) to check a bag, $ 2x for food, and $ (15x - 16) to upgrade to first class.

vagabundo [1.1K]

42x + 1x + 2x + 1x
Hope this helps

8 0

2 years ago

If f(x) = 5x - 7, what is {3)?<br> O A. 22<br> O B. 2<br> O c. 1<br> • D. 8

DerKrebs [107]

<h3>Answer: 8</h3>

Work Shown:

f(x) = 5x - 7

f(3) = 5(3) - 7 .... every x replaced with 3

f(3) = 15 - 7

f(3) = 8

6 0

2 years ago

Read 2 more answers

| x- ( - 18) | if x < - 18

aleksandrvk [35]

Answer:

well x-(-18) x+18 but the if part is x<-18 will stay the same

Step-by-step explanation:

6 0

1 year ago

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

2 years ago