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Paraphin [41]
3 years ago
6

Which of the following probabilities is the greatest for a standard normal distribution?P (negative 1.5 less-than-or-equal-to z

less-than-or-equal-to negative 0.5) P (negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5) P (0.5 less-than-or-equal-to z less-than-or-equal-to 1.5) P (1.5 less-than-or-equal-to z less-than-or-equal-to 2.5)
Mathematics
2 answers:
Darina [25.2K]3 years ago
5 0

Answer:

the answer is below

Step-by-step explanation:

Which of the following probabilities is the greatest for a standard normal distribution?P (negative 1.5 less-than-or-equal-to z less-than-or-equal-to negative 0.5) P (negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5) P (0.5 less-than-or-equal-to z less-than-or-equal-to 1.5) P (1.5 less-than-or-equal-to z less-than-or-equal-to 2.5)

Alinara [238K]3 years ago
4 0

Answer:

negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5

Step-by-step explanation:

its the peak of the graph

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Charra [1.4K]

Answer:

37

Step-by-step explanation:

-3x^{2} + 1

replace (x) with (-2)

-3(-2)^{2} + 1

6^2 +1=36+1

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Which of the value of x satisfies the equation below -2x-7=12
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Answer is A.
You start by adding the 7 to the opposite side. Giving you 19. You divide both sides by -2 which gives you x on one side and -9.5 on the other.

-2x-7=12
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2 years ago
A car is traveling at 70 ​mi/hour. What is the​ car's speed in feet per​ second?
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3 years ago
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Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel
GarryVolchara [31]

Answer:

The unusual X values ​​for this model are: X = 0, 1, 2, 7, 8

Step-by-step explanation:

A binomial random variable X represents the number of successes obtained in a repetition of n Bernoulli-type trials with probability of success p. In this particular case, n = 8, and p = 0.53, therefore, the model is {8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}. So, you have:

P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024

P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215

P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848

P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912

P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695

P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431

P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371

P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442

P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062

The unusual X values ​​for this model are: X = 0, 1, 7, 8

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Step-by-step explanation:

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