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Mrac [35]
3 years ago
5

#41.

Mathematics
1 answer:
yaroslaw [1]3 years ago
7 0
No since the between consecutive numbers doesn't give a constant
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Base = 5 ft Height= 3.5 ft what is the area
Advocard [28]

Answer:

the area is 17.5 ft if the shape is a square/rectangle

the area is 8.75 ft if the shape is a triangle

Step-by-step explanation:

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be X centimeters.

X follows a normal distribution with

  • mean \mu = 11.2 and
  • standard deviation \sigma = 2.1.

In other words,

X\sim \text{N}(11.2, 2.1^{2}),

and the probability in question is X \le 13.

Z-score table approach:

Find the z-score of this measurement:

\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143. Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that Z will be less than or equal to the z-score in question. (In case the question is asking for the probability that Z is greater than the z-score, subtract the value from table from 1.)

P(X\le 13) = P(Z \le 0.857143) \approx 0.8051.

"Technology" Approach

Depending on the manufacturer, the steps generally include:

  • Locate the cumulative probability function (cdf) for normal distributions.
  • Enter the lower and upper bound. The lower bound shall be a very negative number such as -10^{9}. For the upper bound, enter 13
  • Enter the mean and standard deviation (or variance if required).
  • Evaluate.

For example, on a Texas Instruments TI-84, evaluating \text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 ) gives 0.804317.

As a result,

P(X\le 13) = 0.804317.

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

  • Number of trials n: 150.
  • Probability of success p: 0.804317.

Let Y be the number of green beans out of this 150 that are shorter than 13 centimeters. Y\sim\text{B}(150,0.804317).

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

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