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kenny6666 [7]
2 years ago
15

Find five numbers so that the mean, median, mode and range is 4.​

Mathematics
1 answer:
Nuetrik [128]2 years ago
8 0

Answer:

Let’s break this apart

Well we know the median has to be 4

Since it’s 5 numbers the middle number has to be 4 since its the median.

Let’s put in what we know.

a, b, 4, d, e

Constraints:

There has to be more then 1 “4”.

e-a = 4

So using that information lets solve since the possibility is almost endless

<u></u>

SO lets make e 5 and a 1.

1, b, 4, d, 5

There has to be more then 1 4 so lets put that as b.

We can solve for the last remaining digits.

1+ 4 + 4 + d + 5 / 5 = 4

14 + d /5 = 4

2.2 + d = 4

1.8 = d

So now if we put in order and replace b with 1.8 and make d as the previous “b” as 4.

1, 1.8, 4, 4, 5

Thats your 5 numbers right there.

Check:

Mode is 4: yes!

Range is 4: 5-1 = 4 yes!

Median is 4: yes!

Mean is 4: 1 + 1.8 + 4 + 4 +5 / 5 = 4 yes!

Every thing checks out.

There could be a lot of possibilities.

For example take this wrong one

Lets make the same exact thing except change the a and the e.

THis is what we have,

a, b, 4, d, e

Lets make e and a as 6 and 2.  6-2 is still 4 so its possible.

2, b, 4, d, 6

And of course we need more then 1 4 so lets make d 4.

2, b, 4, 4, 6

Now solve for b in the mean.

2 + b + 4 + 4 + 6  / 5 = 4

16 +b /5 = 4

3.2 + b = 4

Solve

B = 0.8

This doesn’t work cause the median and the range has constraint here…

When doing a median, it has to be in ORDER.

2, 0.8, 4, 4 , 6 isn’t in order

ANd even when put in order.

0.8, 2, 4, 4, 6

THe range has the constraint here becuase 6 - 0.8 isn’t 4.

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A. y = -1.33x; 2.67<br> B. y = -0.75x; 1.50<br> C. y = 1.33x; -2.67<br> D. y = 0.13x; -0.25
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3 years ago
The average American man consumes 9.8 grams of sodium each day. Suppose that the sodium consumption of American men is normally
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Answer:

(a) The distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

Step-by-step explanation:

The random variable <em>X</em> is defined as the amount of sodium consumed.

The random variable <em>X</em> has an average value of, <em>μ</em> = 9.8 grams.

The standard deviation of <em>X</em> is, <em>σ</em> = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable <em>X</em> follows a normal distribution with parameters <em>μ</em> = 9.8 grams and <em>σ</em> = 0.8 grams.

Thus, the distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

To compute the probability of  Normal distribution it is better to first convert the raw score (<em>X</em>) to <em>z</em>-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

P(8.8

                           =P(-1.25

Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c)

The probability representing the middle 30% of American men consuming sodium between two weights is:

P(x_{1}

Compute the value of <em>z</em> as follows:

P(-z

The value of <em>z</em> for P (Z < z) = 0.65 is 0.39.

Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:

-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5     z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

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Algebraically determine the points of intersection of (x-1)^2+(y-2)^2=4 and y-2=2x
dlinn [17]
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