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Elena L [17]
3 years ago
8

A normal distribution has a mean of 50 and standard deviation of 5. Which value produces a negative z-score?​

Mathematics
1 answer:
zimovet [89]3 years ago
8 0

Answer:

x\:.

Step-by-step explanation:

The z-score for a normal distribution is calculated using the formula:

Z=\frac{x-\mu}{\sigma}.

From the question, the distribution has a mean of 50.

\implies \mu=50 and the standard deviation is \sigma=5.

For a z-score to be negative, then, \frac{x-\mu}{\sigma}\:.

\frac{x-50}{5}\:.

x-50\:.

x-50\:.

x\:.

\therefore x\:.

Any value less than 50 will produce a negative z-score

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The number of primes less than 190 using the principle of inclusion-exclusion are 42

<h3>Principle of inclusion- exclusion</h3>

The principle of inclusion-exclusion is known as a counting technique that computes the number of elements satisfying at least one of several properties and guaranteeing that the numbers are not counted twice.

Prime numbers are numbers only divisible by 1 and itself.

Prime numbers less than 190 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181

They are 42 in number

Therefore, the number of primes less than 190 using the principle of inclusion-exclusion are 42

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Step 2: Subtract 80 from both sides.

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Step 3: Divide by both sides by 3.

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