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3 years ago

Can anybody tell me the answers to number 1, 2 & 3 ???!!!!

Mathematics

1 answer:

Novay_Z [31]3 years ago

6 0

Problem 1

Domain = {-1, -3, 2, 1}

Range = {5, 0, 2}

The domain is the set of possible inputs and the range is the set of possible outputs. This is a function because each input goes to exactly one output.

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Problem 2

This is a function as well. We do not have any input going to multiple outputs.

Domain = {-2, -3, 5}

Range = {6, 7, 8}

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Problem 3

This is not a function. The input -4 goes to more than one output (outputs 3 and -1 at the same time)

Domain = {-4, -2, 0}

Range = {3, -1, -2, 4}

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The summer monsoon brings 80% of India's rainfall and is essential for the country's agriculture.

Natasha_Volkova [10]

Answer:

Step 1. Between 688 and 1016mm. Step 2. Less than 688mm.

Step-by-step explanation:

The 68-95-99.7 rule roughly states that in a normal distribution 68%, 95% and 99.7% of the values lie within one, two and three standard deviation(s) around the mean. The z-scores represent values from the mean in a standard normal distribution, and they are transformed values from which we can obtain any probability for any normal distribution. This transformation is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ (1)

$\\ \mu\;is\;the\;population\;mean$

$\\ \sigma\;is\;the\;population\;standard\;deviation$

And x is any value which can be transformed to a z-value.

Then, z = 1 and z = -1 represent values for one standard deviation above and below the mean, respectively; values of z = 2 and z =-2, represent values for two standard deviations above and below the mean, respectively and so on.

Because of the 68-95-99.7 rule, we know that approximately 95% of the values for a normal distribution lie between z = -2 and z = 2, that is, two standard deviations below and above the mean as remarked before.

<h3>Step 1: Between what values do the monsoon rains fall in 95% of all years?</h3>

Having all this information above and using equation (1):

$\\ z = \frac{x - \mu}{\sigma}$

For z = -2:

$\\ -2 = \frac{x - 852}{82}$

$\\ -2*82 + 852 = x$

$\\ x_{below} = 688mm$

For z = 2:

$\\ 2 = \frac{x - 852}{82}$

$\\ 2*82 = x - 852$

$\\ 2*82 + 852 = x$

$\\ x_{above} = 1016mm$

Thus, the values for the monsoon rains fall between 688mm and 1016mm for approximately 95% of all years.

<h3>Step 2: How small are the monsoon rains in the driest 2.5% of all years?</h3>

The driest of all years means those with small monsoon rains compare to those with high values for precipitations. The smallest values are below the mean and at the left part of the normal distribution.

As you can see, in the previous question we found that about 95% of the values are between 688mm and 1016mm. The rest of the values represent 5% of the total area of the normal distribution. But, since the normal distribution is symmetrical, one half of the 5% (2.5%) of the remaining values are below the mean, and the other half of the 5% (2.5%) of the remaining values are above the mean. Those represent the smallest 2.5% and the greatest 2.5% values for the normally distributed data corresponding to the monsoon rains.

As a consequence, the value x for the smallest 2.5% of the data is precisely the same at z = -2 (a distance of two standard deviations from the mean), since the symmetry of the normal distribution permits that from the remaining 5%, half of them lie below the mean and the other half above the mean (as we explained in the previous paragraph). We already know that this value is x = 688mm and the smallest monsoons rains of all year are less than this value of x = 688mm, representing the smallest 2.5% of values of the normally distributed data.

The graph below shows these values. The shaded area are 95% of the values, and below 688mm lie the 2.5% of the smallest values.

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Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights

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Answer:

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

Given that, A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171, Z = (4171 - 3311)/860 = 1

P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.

Next, multiply that by the sample size of 732.

Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171

For part b, use the same method except x is now 1591.

Z = (1581 - 3311)/860 = -2

P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

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Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

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