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jok3333 [9.3K]
4 years ago
15

How do theoretical probability and experimental probability differ?

Mathematics
1 answer:
Aleks [24]4 years ago
6 0

Answer:

differentiated by the method of calculating the probability of an event. In experimental probability, the success and the failure of the concerned event are measured/counted in a selected sample and then the probability is calculated.

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given the following system of equations, identify the type of system.x y = 6y = 3 - xconsistentinconsistentequivalent
lapo4ka [179]
Inconstant as there is no solution
8 0
3 years ago
On Monday it snowed 9 inches. The next day it snowed 1/2 that amount. How much did it snow on the second day?
riadik2000 [5.3K]
It probably snowed 4.5 inches
3 0
3 years ago
The Census Bureau says that the 10 most common names in the United States are (in order) Smith, Johnson, Williams, Jones, Brown,
scZoUnD [109]
Using binomial distribution where success is the appearing of any of the top 10 most common names, thus probability of success (p) is 9.6% = 0.096 and the probability of failure = 1 - 0.096 = 0.904. Number of trials is 11.
Binomial distribution probability is given by P(x) = nCx (p)^x (q)^(n - x)
Probability that none of the top 10 most common names appears is P(0) = 11C0 (0.096)^0 (0.904)^(11 - 0) = (0.904)^11 = 0.3295
Thus, the probability that at least one of the 10 most common names appear is 1 - 0.3295 = 0.6705

Therefore, I will be supprised that none of the names of the authors were among the 10 most common names given that the probability that at least one of the names appear is 67%.
6 0
3 years ago
A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measurin
Ratling [72]

Answer:

The standard deviation of weight for this species of cockroaches is 4.62.

Step-by-step explanation:

Given : A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

To find : What is the approximate standard deviation of weight for this species of cockroaches?

Solution :

We have given,

Mean \mu=50

The sample mean x=55

A lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

i.e. P(X>55)=14%=0.14

The total probability needs to sum up to 1,

P(X\leq 55)=1-P(X>55)

P(X\leq 55)=1-0.14

P(X\leq 55)=0.86

The z-score value of 0.86 using z-score table is z=1.08.

Applying z-score formula,

z=\frac{x-\mu}{\sigma}

Where, \sigma is standard deviation

Substitute the values,

z=\frac{x-\mu}{\sigma}

1.08=\frac{55-50}{\sigma}

1.08=\frac{5}{\sigma}

\sigma=\frac{5}{1.08}

\sigma=4.62

The standard deviation of weight for this species of cockroaches is 4.62.

4 0
3 years ago
Can any one help with me this one please
goblinko [34]

Answer:

∠ e = 40°

Step-by-step explanation:

Since Δ ABC and Δ EDC are congruent then corresponding angles are congruent, that is

∠ A = ∠ E = 40°

8 0
3 years ago
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