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atroni [7]
3 years ago
5

If u times 23 by 10 shouldn't u get 230 but my teacher says that's not it u have to did everything but I don't know what to did

die
Mathematics
2 answers:
cricket20 [7]3 years ago
4 0
It is 230 I just clculated
makvit [3.9K]3 years ago
3 0
Um well I think you should double check
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At a large university, the mean amount spent by students for cell phone service is $58.90 per month with a standard deviation of
zhenek [66]

Answer:

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 58.90, \sigma = 3.64, n = 44, s = \frac{3.64}{\sqrt{44}} = 0.54875

What is the probability that the mean amount of their monthly cell phone bills is more than $60?

This is 1 subtracted by the pvalue of Z when X = 60. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{60 - 58.90}{0.54875}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

4 0
4 years ago
Given the system of equations, what is the y-coordinate of the solution?5x - 4y = 7x = 5 - y
sveticcg [70]
5x - 4y = 7
x = 5 - y

5(5 - y) - 4y = 7
25 - 5y - 4y = 7
-5y - 4y = 7 - 25
-9y = - 18
y = -18/-9
y = 2 <== y coordinate is 2

6 0
3 years ago
Determine whether the following value is a continuous random variable, discrete random variable, or not a random variable.
Elina [12.6K]

Answer:

a. The number of light bulbs that burn out in the next year in a room with 19 bulbs: is a discrete random variable.

b. The usual mode of transportation of people in City Upper A: is not a random variable because its outcome isn't numerical.

c. The number of statistics students now doing their homework: is a discrete random variable.

d. The number of home runs in a baseball game: is a discrete random variable.

e. The exact time it takes to evaluate 67 plus 29: is a continuous random variable.

f. The height of a randomly selected person: is a continuous random variable.

Step-by-step explanation:

A random variable often used in statistics and probability, is a variable that has its possible values as numerical outcomes of a random experiment or phenomenon. It is usually denoted by a capital letter, such as X.

In statistics and probability, random variables are either continuous or discrete.

1. A continuous random variable is a variable that has its possible values as an infinite value, meaning it cannot be counted.

Example are the height of a randomly selected person, time it take to move from Texas to New York city, etc.

2. A discrete random variable is a variable that has its possible values as a finite value, meaning it can be counted.

Examples are the number of light bulbs that burn out in the next year in a room with 19 bulbs, the number of chicken in a district etc.

4 0
3 years ago
Read 2 more answers
Create two real world problems that involve the fraction 1/2 and the whole number 3
kompoz [17]
You order 4 pizzas for a party. At the end of the party only half of a pizza is left. How much pizza did the guests eat? That's one problem can't think of a 2nd one.
8 0
3 years ago
Consider randomly selecting a single individual and having that person test drive 3 different vehicles. Define events A1, A2, an
zmey [24]

a. Use the inclusion/exclusion principle.

P(A_1\cap A_2)=P(A_1)+P(A_2)-P(A_1\cup A_2)=0.55+0.65-0.80=0.40

b. By definition of conditional probability,

P(A_2\mid A_3)=\dfrac{P(A_2\cap A_3)}{P(A_3)}=\dfrac{0.50}{0.70}\approx0.7143

4 0
3 years ago
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