What is 850% as a fraction in simplest form
2 answers:
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You’re trying to find the better buy so you want to find out how much you’re paying per unit (in this case kg)
So you divide the price by the unit (kg)
And find the better buy by looking at which price is cheapest.
Your answer is the 7.48 kg bag
Hope this helps and good luck!!
So you divide the price by the unit (kg)
And find the better buy by looking at which price is cheapest.
Your answer is the 7.48 kg bag
Hope this helps and good luck!!
Answer:
a)
b) For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%
And then the % above would be 100-84= 16%
Step-by-step explanation:
Assuming this question : "Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly normal distribution with mean 336 days and standard deviation 6 days. Use the 68-95-99.7 rule to answer the following questions. "
(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?
First we need to remember the concept of empirical rule.
From this case we assume that where X represent the random variable "length of horse pregnancies from conception to birth"
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean so then we can find the limits for this case with this:
(b) What percent of horse pregnancies are longer than 342 days?
For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%
And then the % above would be 100-84= 16%