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Deffense [45]
3 years ago
15

Can x^3 be continuous, or is it discrete?

Mathematics
1 answer:
White raven [17]3 years ago
4 0
A continuous random variable (X) has an infinite number of values within an interval whereas a discrete random variable (X) assumes a value among a finite set including X1, X2, X3, and so on.
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Determine the number of solutions for the equation shown below
Mila [183]
The answer would be B
5 0
3 years ago
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2= d-11/3 <br><br> 100 points if u answer pls help asap
hjlf

Answer:

i hope this works

Step-by-step explanation:

d=17/3

d=5.666

d=5 2/3

8 0
3 years ago
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What is the grade average of 75 70 90 60 50 90 70
Iteru [2.4K]
The average can be found by adding them all up and dividing that number by the number of items. If the result have you a number that's more than the largest or less than the lowest, you did it wrong.

(75+70+90+60+50+90+70)/7
= 505/7
=72.14
6 0
4 years ago
Currently, I am $9$ times as old as my son. Next year, I will be $7$ times as old as my son. How old is my son now?
grandymaker [24]

Answer:

He is 3 years old.

Step-by-step explanation:

3 times 9 is 27. She is 9 times older than her 3 year old son.

Next year he will be 4 and she will be 28. 4 times 7 is 28. She will be 7 times older than her son.

6 0
3 years ago
Read 2 more answers
A. Use the​ one-mean t-interval procedure with the sample​ mean, sample​ size, sample standard​ deviation, and confidence level
PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668

Margin of error, E = t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

M.E = \frac{22.773 - 17.227}{2} = 2.773

c)M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}

= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

4 0
3 years ago
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