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Nina [5.8K]
3 years ago
11

She mows 1/6 acres in 1/4 hour .? How many acres she mows per hour

Mathematics
1 answer:
Vlad [161]3 years ago
8 0
1/6 acres in 1/4 hour
1/6 * 4 = 4/6 = 2/3
she mows 2/3 acres per hour
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EASY MATH 10 PTS + BRAINLIEST
mina [271]

Answer:

d. 5

Step-by-step explanation:

4 0
3 years ago
Solve this I'll give brainliest plz look at the picture ​
Diano4ka-milaya [45]

Answer:

answer is B

Step-by-step explanation:

the number that 10% of 20 is coming from 200 so 23% of 200 is 46

8 0
2 years 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
3 years ago
Dion practiced the long jump . He jumped 3.05 meters,2.74 meters and 3.3 meters.What was the average length of Dion's jumps?
Mademuasel [1]
All you do is subtract 3.05 and 2.74 then do the same thing with the other ones and see what number is between all of them. What grade are you in
5 0
3 years ago
Read 2 more answers
How to solve (x-10)(x + 10)
tensa zangetsu [6.8K]

Answer:

Multiplying/simplifying, it is x^2-100(found by the FOIL method)

Step-by-step explanation:

multiply the commons factors x times x is x squared

-10 times 10 is -100. Negative times positive is negative.

6 0
3 years ago
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