What is 12/6 equivalent to

Mathematics

2 answers:

Bond [772]3 years ago

4 0

12/6 is equivalent to 2.

hope this helps!

sveta [45]3 years ago

3 0

Ahh, yes, college can be very had for math. I do struggle with simple fractions.

Jokes aside, 12/6 can be simplified to 2/1 or plainly 2

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Can someone please help?

REY [17]

Answer:

(a) no

(b) -1, (2, -1)

(d) no

Step-by-step explanation:

You can check by using x=3 in the function.

f(3) = 4(3) -9 = 12 -9 = 3 . . . . not -3

The point (3, -3) is not on the graph.

Put x=2 into the function and evaluate:

f(2) = 4(2) -9 = 8 -9

f(2) = -1

The point (2, -1) is on the graph.

Put f(x) = 7 into the equation and solve for x.

7 = 4x -9

16 = 4x

4 = x

The point (4, 7) is on the graph.

no, see part (b)

4 0

2 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .