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3 years ago

Find the Average Number of Fruits eaten By Tony on a week:

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

7 0

Answer:

41.43 fruits

Step-by-step explanation:

Average number of fruits= sum of fruits for each dat ÷ total number of fruits

$\frac{60 + 40 + 20 + 40 + 80 + 30 + 20}{7}$

$= \frac{290}{7}$

$= 41.43 \: fruits$

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What are the chances of landing on heads when tossing one coin?

e-lub [12.9K]

Answer:

50%

Step-by-step explanation:

Since there is either heads or tails, meaning only two options you have a 50% chance of getting heads

6 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

What's the easiest way to answer how I know the answer pls?

-BARSIC- [3]

Answer: Table C

Explanation: The X values match up with those on the graph!

4 0

2 years ago

A right square pyramid has a slant height of 10 feet, and the length of a side

kotykmax [81]

Step-by-step explanation:

take the base square. find the diagonal

16^2+16^2 = 2*16^2= d^2

d= 16√2

half of the diagonal will form the baSe of right triangle of hypotenuse 10. but the base will be 8√2 which is greater than 10. so I feel the slant hr has to be different and more than 10

6 0

2 years ago

A help would be greatly appreciated!

dalvyx [7]

The tree was approximately 40.8ft tall.

You use the equation, a^2+b^2=c^2.
12^2 + 39^2 = x^2, add the results of 12^2 and 39^2 to make 1665 = x^2. You can take the positive root of both sides, 3√185, and it equals 40.80441152, which is rounded to 40.8.

7 0

3 years ago