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KengaRu [80]
3 years ago
14

What is the sum of 15 and its additive inverse?

Mathematics
1 answer:
vazorg [7]3 years ago
5 0

Answer:

O C. 0

Step-by-step explanation:

What is the sum of 15 and its additive inverse?

In Mathematics, the additive inverse of a number is defined as the number that when it is added to a specific number equals 0

From the above question, the additive of 15 = -15

Therefore, the sum of 15 and its additive inverse is calculated as:

15 + -15

= 15 - 15

= 0

Therefore, Option C is the correct option

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if U={natural number less than 10}, M={multiples of 2}, N={factors of 8} and O={even number} then can we write M intersection (N
mr_godi [17]

Answer:

yes

Step-by-step explanation:

u={1,2,3,4,5,6,7,8,9}

m={2,4,6,8}

N={1,2,4,8}

O={2,4,6,8}

Now

M intersection (N intersection O) = (M intersection N) intersection O

or , {2,4,6,8} intersection {2,4,8} ={2,4,8}

intersection {2,4,6,8 }

:. {2,4,8} ={2,4,8}

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3 years ago
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

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3 years ago
Who good at angles ?
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Answer:

JRQ = 56

Step-by-step explanation:

SRQ - SRJ = JRQ

166 - 110 = 56

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Sam's depth is 25 feet as 14 + 9 = 25

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