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Rashid [163]
3 years ago
9

Which expressions are equivalent to 5(-2k-3)+2k

Mathematics
1 answer:
bulgar [2K]3 years ago
6 0

Answer:

-8k - 15

Step-by-step explanation:

5(-2k-3)+2k

At first, we will break the parenthesis. To break that, we will multiply the value inside the parenthesis by the adjacent number, that is 5. Again, we have to consider the Algebraic operation (Golden rule) -

[(-) x (-) = (+); (+) x (-) = (-)]

Therefore, since there is a minus sign in each of the value inside the parenthesis, the result will be minus as 5 is a positive integer.

or, -5*(2k) - (5*3) + 2k

or, -10k - 15 + 2k

or, -8k - 15 (after the deduction)

The answer is = -8k - 15

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For a certain experiment, if we expect to get a red marble twelve times in 60 trials, what is the theoretical probability of get
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Read 2 more answers
A random sample is selected from a population with mean μ = 102 and standard deviation σ = 10. For which of the sample sizes wou
chubhunter [2.5K]

Answer:

Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the \bar X with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.

\bar X \sim N (\mu ,\frac{\sigma}{\sqrt{n}} )

Based on this rule we can conclude:

a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440

Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean

for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed

Step-by-step explanation:

For this case we know that for a random variable X we have the following parameters given:

\mu = 102, \sigma =10

Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the \bar X with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.

\bar X \sim N (\mu ,\frac{\sigma}{\sqrt{n}} )

Based on this rule we can conclude:

a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440

Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean

for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed

8 0
3 years ago
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