Answer:
Step-by-step explanation:
Question 1. 50% of what number is 100?
Number = x
50%x = 100
5x/10 = 100 (50% = 2/10)
5x = 1000 (Multiplying by 10 at both sides of the equation)
<u>x = 200</u> (Dividing by 5 at both sides of the equation)
<u>Part 50 Whole 100 Part 75 Whole 150 Part 100 Whole 200</u>
Question 2. 25% of what number is 30?
Number = x
25%x = 30
x/4 = 30 (25% = 1/4)
x = 120 (Multiplying by 4 at both sides of the equation)
<u>Part 25 Whole 30 Part 50 Whole 60 Part 75 Whole 90 Part 100 Whole 120</u>
<u>Question 3. What did you differently to find the whole using a ratio table?</u>
Using a ratio table is a much more simple path to get the whole. I take it and just calculate the double of the information given to me for completing the next values in my table, until I get the part 100 and its correspondent whole.
60 months is 5 years
12 × 5 = 60
Interest = Principle amount × Rate × Time in years
= 650 × 0.06 × 5 (6% =
![\frac{6}{100}](https://tex.z-dn.net/?f=%20%5Cfrac%7B6%7D%7B100%7D%20)
= 0.06)
= R195
R650 + R195 =
R845
Answer:
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and
From the central limit theorem we know that the distribution for the sample mean
is given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
Part a
The mean is ![\mu_{\bar x }= 17](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%20%7D%3D%2017)
Part b
And the deviation:
![\sigma_{\bar X}= \frac{5.6}{\sqrt{13}}= 1.553](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B5.6%7D%7B%5Csqrt%7B13%7D%7D%3D%201.553)
Step-by-step explanation:
Assuming this complete info: Suppose a random variable xx is normally distributed with μ=17 and σ=5.6. According to the Central Limit Theorem, for samples of size 13:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and
From the central limit theorem we know that the distribution for the sample mean
is given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
Part a
The mean is ![\mu_{\bar x }= 17](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%20%7D%3D%2017)
Part b
And the deviation:
![\sigma_{\bar X}= \frac{5.6}{\sqrt{13}}= 1.553](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B5.6%7D%7B%5Csqrt%7B13%7D%7D%3D%201.553)
Answer:
x>3
Step-by-step explanation:
treat the inequality like an equal sign