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ASHA 777 [7]
3 years ago
15

A scientist reviews her data to find the measures of central tendency using a calculator. She finds the average mass of an adult

brown bear is about 1.36E5 grams, and the average mass of a beetle fly is approximately 0.61E2 grams. She also found the sum total of her data using a calculator. The total brown bear population in the world is approximately 2E5, and the total beetle fly population is approximately 7E10. Rounded to the nearest whole number, the total mass of all living beetle flies is approximately times more than the total mass of all living brown bears.
answer is 157


First, find the average mass of an adult brown bear. The average mass is about 1.36E5 grams which is equivalent to 1.36 × 105 grams in scientific notation.

The total brown bear population in the world is approximately 2E5 which is equivalent to 2 × 105 in scientific notation.

So, to find the total mass, multiply the average mass of an adult brown bear and the total number of brown bear population.



Therefore, the total mass of the brown bear population is 2.72 × 1010 grams.

Next, find the average mass of a beetle fly. The average mass is about 0.61E2 grams which is equivalent to 0.61 × 102 grams in scientific notation.

The total beetle fly population is approximately 7E10 which is equivalent to 7 × 1010 in scientific notation.

So, to find the total mass, multiply the average mass of beetle fly and the total number of beetle fly population.



Therefore, the total mass of the beetle fly population is 4.27 × 1012 grams.
Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

Hello

Step-by-step explanation:

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7y

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3 0
3 years ago
Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636​, the central limit theorem appl
lakkis [162]

Answer:

a) \bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1

b) P(\bar X >101)=1-P(\bar X

c) P(\bar X

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 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".  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

Let X the random variable that represent the variable of interest on this case, and for this case we know the distribution for X is given by:  

X \sim N(\mu=100,\sigma=6)  

And let \bar X represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:  

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})  

a. What are the mean and variance of the sampling distribution for the sample​ means?

\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1.2^2=1.44

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:  

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

If we apply this formula to our probability we got this:  

z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833  

And we want to find this probability:

P(\bar X >101)=1-P(\bar X

On this last step we use the complement rule.  

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67  

And we want to find this probability:

P(\bar X

5 0
4 years ago
F(x) = 12 + x<br> X<br> f(x)<br> -3<br> -1<br> 0<br> 4<br> 6
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9514 1404 393

Answer:

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Step-by-step explanation:

The function definition tells you that adding 12 to the x-value will give you the value of f(x).

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The (x, f(x)) values for the table are shown above.

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