64-144a+108a^2-27a^3

You and a friend have a hula hoop competition. you win 3 out of every 7 trials. your friend wins 5 more trials than you. how man

Ivenika [448]

Answer:

Total number of trials is 35.

Step-by-step explanation:

Given : Me and my friend have a hula hoop competition. I win 3 out of every 7 trials. My friend wins 5 more trials than me.

To find : How many trials are in the competition?

Solution :

Probability of my winning is 3 out of 7 trials $=\frac{3}{7}$

and probability of my friend's winning is $=1-\frac{3}{7}=\frac{4}{7}$

Let there be n numbers of trials.

Therefore, my score after n trials $=\frac{3n}{7}$

and my friend's score after n trials $=\frac{4n}{7}$

Now, the number of trials my friend won more than me is

$\frac{4n}{7}-\frac{3n}{7}=\frac{n}{7}$

We know, that my friend won 5 more trials than me so comparing the actual and given we get,

$\frac{n}{7}=5\\n=7\times5=35$

Therefore, total number of trials is 35.

We can also verify our answer by placing value of n,

After 35 trials I will win $\frac{3\times 35}{7}=15$ trials

and my friend will win $\frac{4\times 35}{7}=20$ trials.

Hence, my friend won 5 more trials than me.

3 0

2 years ago

As part of a community service project at your high school, you are organizing a fund-raiser at the neighborhood movie theater.

Artist 52 [7]

The answer is
A.
Independent variable - number of persons attending (or number of tickets sold);
Dependent variable - the amount donated

Because, the number of persons attending is independent. No one knows how many people are attending. But the amount donated is depended upon the number of people attending.

6 0

2 years ago

in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru

melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is

n = 200

the proportion is

p = 154/200 = 0.77

Since both np ≥ 10 and n(1-p) ≥ 10

We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference.

The approximation would be to a Normal curve with this parameters:

Mean

p = 0.77

Standard deviation

$\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298$

The 90% confidence interval for the proportion would be then

$\bf [0.77-z^*0.0298, 0.77+z^*0.0298]$

where $\bf z^*$ is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval $\bf [-z^*,z^*]$ is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

NORMSINV(0.95) and we get a value of 1.645

The 90% confidence interval for the proportion is then

$\bf [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]$

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0

2 years ago

Which rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C'?

kirill [66]

Answer:

From the figure, it can be seen that the mage A'B'C' is obtained from the pre-image ABC by translating/shifting the vertices of the image by 7 units to the right and 6 units down.

Therefore, the rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C' is (x, y) → (x + 7, y – 6).

8 0

1 year ago

In which quadrants is the ordinate positive?

Anni [7]

Answer:

I and II

Step-by-step explanation:

The ordinate is the y element in an ordered pair in a Cartesian coordinate system. The quadrants in which the ordinates are positive should be at quadrants I and II.

7 0

1 year ago

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