Y=-1/2-16K+10 in vertex form

A probability experiment is conducted in which the sample space of the experiment is upper s equals { 1 comma 2 comma 3 comma 4

Softa [21]

You can think of this experiment as follows: you have to select a random number between 1 and 15, where each number is equally likely to appear.

Player A wins if 3,4,5 or 6 are picked.

Player B wind is 10,11 or 12 are picked.

Two events are said to be mutually exclusive if they cannot happen at the same time. So, in our metaphor, the question is: can we pick a particular number, so that both players win simultaneously?

The question is no, because if player A wins, we have picked 3,4,5 or 6, and player B wins with none of these numbers.

Similarly, if player B wins, we have picked 10, 11 or 12, and player A wins with none of these numbers.

So, the two events cannot occur simultaneously.

If you want to be more formal, since the events are subset of the sample space, two events are mutually exclusive if their intersection is the empty set, and this is the case:

$A \cap B = \{3,4,5,6\}\cap\{10,11,12\} = \emptyset$

8 0

2 years ago

What conclusions can be drawn using the given and its diagram?

dem82 [27]

Answer:

Whichever reasoning processes and research methods were used, the final conclusion is critical, determining success or failure. If an otherwise excellent experiment is summarized by a weak conclusion, the results will not be taken seriously.

Success or failure is not a measure of whether a hypothesis is accepted or refuted, because both results still advance scientific knowledge.

Failure lies in poor experimental design, or flaws in the reasoning processes, which invalidate the results. As long as the research process is robust and well designed, then the findings are sound, and the process of drawing conclusions begins.Whichever reasoning processes and research methods were used, the final conclusion is critical, determining success or failure. If an otherwise excellent experiment is summarized by a weak conclusion, the results will not be taken seriously.

Success or failure is not a measure of whether a hypothesis is accepted or refuted, because both results still advance scientific knowledge.

Failure lies in poor experimental design, or flaws in the reasoning processes, which invalidate the results. As long as the research process is robust and well designed, then the findings are sound, and the process of drawing conclusions begins.

little info

Step-by-step explanation:

hope this help

pick me as the brainliest

8 0

2 years ago

Last year, nine employees of an electronics company retired. Their ages at retirement are listed below.

GuDViN [60]

Answer:

58.0

Step-by-step explanation:

Given the data:

56 65 62 53 68 58 65 52 56

Reorderd data: 52, 53, 56, 56, 58, 62, 65, 65, 68

Median = 1/2(n + 1) th term

n = sample size = 9

Hence,

Median = 1/2(9 + 1)th term

Median = 1/2(10)th term

Median = 5th term

5th term in the reordered data = 58

Hence, median age is 58

7 0

3 years ago

What is the solution set to the inequality 5(x – 2)(x + 4) > 0?

ankoles [38]

Answer:

Step-by-step explanation:

The Answer is {x| x <-4 or x> 2}

8 0

2 years ago

Read 2 more answers

Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. U

kogti [31]

Answer:

11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Forecast of rain.

Event B: Raining.

In recent years, it has rained only 5 days each year.

A year has 365 days. So

$P(B) = \frac{5}{365} = 0.0137$

When it actually rains, the weatherman correctly forecasts rain 90% of the time.

This means that $P(A|B) = 0.9$

Probability of forecast of rain:

90% of 0.0137(forecast and rains)

10% of 1 - 0.0137 = 0.9863(forecast, but does not rain)

$P(A) = 0.0137*0.9 + 0.9863*0.1 = 0.11096$

What is the probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

$P(B|A) = \frac{0.0137*0.9}{0.11096} = 0.1111$

11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

3 0

2 years ago