Is 2/5 less then 7/15

Mr Smith has 50 papers to grade. He graded 38 papers and he had a scholar grade the rest what percent of the papers did each per

Jet001 [13]

First you will divide 100 by 50 you get 2. then you subrtact 38 from 50. you get 12. then since 100 is twice the amount as 50, you will multiply 12 by 2. you get 24. that is your answer

4 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

hoa [83]

Answer:

Obese people

$Lower = \mu - 2\sigma = 373- 2(67) = 239$

$Upper = \mu + 2\sigma = 373+ 2(67) = 507$

Lean People

$Lower = \mu - 2\sigma = 526- 2(107) = 312$

$Upper = \mu + 2\sigma = 526+ 2(107) = 740$

Step-by-step explanation:

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

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)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

$X \sim N(373,67)$

Where $\mu=373$ and $\sigma=67$

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

$Lower = \mu - 2\sigma = 373- 2(67) = 239$

$Upper = \mu + 2\sigma = 373+ 2(67) = 507$

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

$X \sim N(526,107)$

Where $\mu=526$ and $\sigma=107$

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

$Lower = \mu - 2\sigma = 526- 2(107) = 312$

$Upper = \mu + 2\sigma = 526+ 2(107) = 740$

The interval for the lean people is significantly higher than the interval for the obese people.

5 0

2 years ago

Lara has three cats: Socks, Princess, and Luna. The ratio of Sock's weight to Princess's weight is 4:5. The ratio of Princess's

romanna [79]

A = Sock's weight
b = Princess' weight
c = Luna's weight

a : b = 4 : 5
b : c = 6 : 7

at this point, we have to change the 2 ratios (a : b and b : c) into one ratio (a : b : c)
to do that, we have to change the ratio number so the number b in both ratio is the same

a : b = 4 : 5 = 24 : 30
b : c = 6 : 7 = 30 : 35
From that we can conclude that
a : b : c = 24 : 30 : 35

Since we're looking for the ratio of Sock's weight to Luna's weight, it will be
a : c = 24 : 35

6 0

2 years ago

Read 2 more answers

22. Find the perimeter and area.

zmey [24]

Answer:

Part 22) The area is $A=15a^3b^6\ units^2($ and the perimeter is $P=10a^2b^4+6ab^2\ unit$