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nasty-shy [4]
3 years ago
15

F(t)=3t^2-2 , find f(k-2)

Mathematics
1 answer:
AysviL [449]3 years ago
7 0
3(k-2)² -2
3(k²-4k+4) -2
3k²-12k+12-2
3k² -12k +10
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What is the value of x in the equation −6 + x = −1?
zhenek [66]

Answer:

the answer is d 5

Step-by-step explanation:-6 +5= -1

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I have 1 hundreds, 4 ones, 2 tens, and 1 tenths. What number am I?
DaniilM [7]
Since it reads 1 hundreds, that means 100. 4 ones mean 4 * 1, which is 4. 2 tens means 2 * 10, which is 20. Add those together and you get 124. Where it reads "1 tenth", that means 0.1
Therefore, the number is 124.1 
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3 years ago
A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using 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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

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3 years ago
Solve the compound inequality.<br><br> 9 – 4x ≥ 5 or 4(–1 + x) – 6 ≥ 2
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X ≤ 1 and x ≥ 3
Hope this helps :)
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An expression is shown below: r50 r2 which statement is true about the expression? it is irrational and equal to 2r13 it is irra
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I think it is rational and equal to 6
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