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Reika [66]
3 years ago
8

Jackson's age was 2/5 of the age he will be 20 years from now 7 years ago. how old is jackson now

Mathematics
1 answer:
Dafna11 [192]3 years ago
8 0
1 year old 5 divided by 20 times 2 minus 7
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A company compiles data on a variety of issues in education. In 2004 the company reported that the national college​ freshman-to
nasty-shy [4]

Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as 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)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

8 0
3 years ago
The box plots show the high temperatures in June and August for Denver in degrees Fahrenheit.
Kryger [21]

Hi There!

---------------------------------------------------------

Full Question:

The box plots show the high temperatures in June and August for Denver in degrees Fahrenheit.

Which can you tell about the mean temperatures for these two months?

There is not enough information to determine the mean temperatures.

The mean temperature for August is higher than June's mean temperature.

The mean temperature for June is equal to the mean temperature for August.

The high interquartile range for August pulls the mean temperature above June's mean temperature.

---------------------------------------------------------

Interquartile Range Formula: Q3 - Q1

Interquartile Range for August: 10

Interquartile Range for June: 8

---------------------------------------------------------

Median = Mean

June: 82

August: 82

---------------------------------------------------------

Answer: The mean temperature for June is equal to the mean temperature for August.

---------------------------------------------------------

Hope This Helps :)

5 0
3 years ago
One pack of sliced cheese weighs 350 grams. How many packages would you need to
Semenov [28]

Answer:

3 packages

Step-by-step explanation:

350+350=700

700+350=1100

1000 grams = 1 kilogram

4 0
3 years ago
Please answer this to the best of your ability
lilavasa [31]

Answer:

C

Step-by-step explanation:

5 0
3 years ago
In the Ambrose family, the ages of the three children are three consecutive even integers. If the age of the youngest child is r
gregori [183]

Answer:

X+7

Step-by-step explanation:

If the youngest child's age expression is X+3 then the middle child's would be x+5 and the oldest would X+7. Even numbers have a difference of 2 between them and that is why 2 is gradually added to each age expression. Hope that helps.

6 0
3 years ago
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