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2 years ago

Consider the function RX) = 3x + 1 and the graph of the function g(x) shown below.

Mathematics

1 answer:

aleksandrvk [35]2 years ago

5 0

Step-by-step explanation:

f(x) goes through the point (0, 1).

g(x) goes through (2, 1) for the same y value.

that means g(x) is f(x) translated 2 units to the right.

but we could also look at the same x value (0).

then g(x) goes through (0, -5).

and that means g(x) is f(x) translated 6 units down.

since it is a line function, shifting it left/right or up/down makes no difference to the shape or position of the new curve itself.

as it looks like you don't have 6 in the drop-down menu, then 2 units (to the right) is the desired correct result.

that means g(x) = f(x-2) = 3(x-2) + 1 = 3x - 6 + 1 = 3x - 5

and that is what the graph is showing us.

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3 years ago

The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability

amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

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

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565$

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

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

$Z = \frac{70.5 - 69}{1.97565}$

$Z = 0.76$

$Z = 0.76$ has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

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Hello There!

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