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lisov135
1 month ago
5

you draw a marble from a bag of numbered marbles whose numbers are normally distributed for some reason. your marble is labeled

'27', and because you're psychic, you know that the mean of the numbered marbles is '21' and also that 43.32% of the marbles are between 21 and 27. What is the standard deviation of the bag of marble numbers? (the answer is an integer)
Mathematics
1 answer:
babunello [11.8K]1 month ago
3 0
Let X represent the numbered marbles.

P(21≤ X ≤27) = 0.4332


\frac{P(21-21)}{\sigma_{x}}\leq \frac{X-21}{\sigma_{x}}\leq \frac{27-21}{\sigma_{x}}


P( 0 ≤ z ≤ 6/σx) = 0.4332; for this case, Z stands for the z-score


1.5 = 6/σx


solving for σx gives us 4





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The yearly cost in dollars, y, for a member at a video game arcade based on total game tokens purchased, x, is y =y equals Start
Leona [12618]

Response:

y = \frac{1}{5}x

Step-by-step explanation:

Given data:

Annual cost for members: y = \frac{1}{10}x + 6

Required:

Calculate the yearly expense for non-members

From the context, we infer that:

A non-member pays $0.20 per game;

Annual cost can be determined using the formula;

y = Amount * Number of Games

Where:

Amount = $0.2, Number of Games = x

y = 0.20 * x

Convert 0.2 to a fraction:

y = \frac{2}{10} * x

Divide the numerator and denominator by 2:

y = \frac{1}{5} * x

y = \frac{1}{5}x

Therefore, the yearly expense for non-members is

y = \frac{1}{5}x

6 0
3 months ago
Read 2 more answers
"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight
AnnZ [12381]

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

</pfor>
6 0
3 months ago
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