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mojhsa [17]
3 years ago
6

Javier and his study group designed a word problem, equation, table, and graph that were all supposed to represent the same info

rmation, as shown below.
Word Problem

A movie theater charges $8.00 for each ticket. What is the total cost, y, for x tickets?

Equation

y = 8x

Graph

Movie Theater Costs
A graph with number of tickets sold on the x-axis and total cost on the y-axis. A line goes through points (8, 1), (16, 2).

Table

A 2-column table with 4 rows. Column 1 is labeled x with entries 5, 8, 12, 13. Column 2 is labeled y with entries 40, 64, 96, 104.

Which piece of information does not represent the same information as the other three parts?
word problem
equation
graph
table
Mathematics
1 answer:
kow [346]3 years ago
3 0

Graph is your answer look below for proof

have a good day!

                                                                     

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Algebraic expression for 8+6q+q
irina1246 [14]
All you can do to this expression is simplify.

You need to combine all "like terms." There are two q terms so you need to combine those. To combine "like terms", simply add their coefficients. 6q has a coefficient of 6 and q has a coefficient of 1. So:

6q + 1q = 7q

Therefore your end result is:

8 +7q
3 0
4 years ago
Select the most accurate statement regarding the normal distribution. Group of answer choices It is always the appropriate distr
PSYCHO15rus [73]

Answer:

There is a 95% chance that values will be within ±2 standard deviations of the mean.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this question:

According to the empirical rule, 95% of the measures are within 2 standard deviations of the mean. So the correct statement is:

There is a 95% chance that values will be within ±2 standard deviations of the mean.

4 0
3 years ago
The diagram below shows 5 identical bowls stacked one inside the other. The height of 1 bowl is 2 inches. The height of 5 bowls
padilas [110]

(Bowls, Height) (1, 2)  (5,5)

Slope is (5-2)/(5-1) = 3/4 inch
y = (3/4)x + b
(2) = (3/4)(1) + b
(2)-(3/4) = b

B=1.25.  Y= 0.75*x + 1.25. 

Part B
X is the number of bowls in the stack and Y is the corresponding height of the stack.

 

7 0
3 years ago
(27y^4-3y^3) /(3y) =
Rus_ich [418]
9y^3-y^2 is a required answer
7 0
3 years ago
In a survey of 1000 eligible voters selected at random, it was found that 100 had a college degree. Additionally, it was found t
o-na [289]

Answer:

A. 8%

B. 39.6%

C. 58.4%

D. 41.6%

Step-by-step explanation:

Computation to determine the probability of eligible voter selected at random

First step is to Draw up a contingincy table which will include Rows = Degree/No degree

and Columns= Vote/Not vote

..............Vote..No vote

Degree 80...20...100

(80%*100=80)

(100-80=20)

No Degree 504..396..900

(1000-100=900)

(56%*900=504)

(504-900=396

Totals 584..416...1000

(80+504=584)

(20+396=416)

(900+100=1,000)

Summary

..............Vote..No vote

Degree 80...20...100

No Degree 504..396..900

Total Totals 584..416...1000

A. Calculation to determine the probability of The voter had a college degree and voted in the last presidential election.

P = 80/1,000

P=0.08*100

P=8%

Therefore the probability of The voter had a college degree and voted in the last presidential election will be 8%

B. Calculation to determine the probability of The voter did not have a college degree and did not vote in the last presidential election.

P =396/1000

P=0.396*100

P=39.6%

Therefore the probability of The voter did not have a college degree and did not vote in the last presidential election will be 39.6%

C. Calculation to determine the probability if The voter voted in the last presidential election.

P = 584/1,000

P=0.584*100

P=58.4%

Therefore the probability if The voter voted in the last presidential election will be 58.4%

D. Calculation to determine the probability if The voter did not vote in the last presidential election.

P = 416/1000

P=0.416*100

P=41.6%

Therefore the probability if The voter did not vote in the last presidential election will be 41.6%

8 0
3 years ago
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