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vitfil [10]
3 years ago
5

Levi has joined the a Frequent Coffee Buyer program with a local cafe. The program costs $5 per month which allows him to get a

discounted coffee price of $3 per cup. Answer the questions below regarding the relationship between the of number of cups of coffee purchased and the total monthly cost.
The independent variable, x, represents the
total monthly cost
, and the dependent variable is the
number of coffee cups purchased
, because the
total monthly cost
depends on the
number of coffee cups purchased
.
A function relating these variables is A(x) =A(x)=
.
So A(12) =A(12)=
, meaning 1212
cups of coffee will cost


dollars monthly
.
Mathematics
1 answer:
pentagon [3]3 years ago
4 0

Answer:

8$

Step-by-step explanation:

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Can someone help me with this question please
valentinak56 [21]

The base is the length of the bottom, from x=-4 to x=2, or 6.


The height is the change in y, from y=5 at the top to y=1 at the bottom, so 4.


Area is base times height, 6 × 4 = 24


24


Second choice



3 0
4 years ago
A cereal box has a length of 8 inches, a width of 2 inches and a height of 12 inches. What is the volume of the cereal box?
DochEvi [55]

Answer:

192 in cubed

Step-by-step explanation:

Multiply 12*2= 24*8

6 0
3 years ago
Read 2 more answers
X^3-6x^2 11x-6=0 <br> a. 1,-5,4 1/3 <br> b. 1,5,4 1/3 <br> c. 0,-5,-4 1/3 <br> d. 0,5,-4 1/3
Ksju [112]
X^3 - 6x^2 + 11x - 6 = 0
f(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0
This means that x - 1 is a factor of x^3 - 6x^2 + 11x - 6 = 0

                     x^2 - 5x + 6
                ___________________
x - 1        |     x^3 - 6x^2 + 11x - 6
             - |     x^3 - x^2
               |______________     
               |          -5x^2 + 11x - 6  
             - |          -5x^2 + 5x
               |_____________________
               |                        6x - 6
             - |                        6x - 6

x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)
Therefore, solutions to x^3 - 6x^2 + 11x - 6 are 1, 2 and 3.
4 0
4 years ago
the seventh grade class is putting on a variety show to raise money. it cost 700 dollars to rent the banquet hall that they are
nexus9112 [7]

Answer:

Since the question is partial, i will assume 2 scenarios:

They need to raise 1000 income

They need to make 1000 as profit

If $1000 as income:

Each ticket costs $15, so  tickets would bring them $1000 income. Fractional ticket is not possible, so rounding gives us 67 tickets as the answer.

If $1000 as profit:

Their cost of renting is $700. We know that .

So, . So, to raise $1700, we need  tickets. Fractional ticket is not possible, so rounding gives us 114 tickets as the answer.

ANSWER:

If need to raise atleast $1000 as income, they need to sell 67 tickets.

If need to raise atleast $1000 as profit, they need to sell 114 tickets.

The most probable answer would be 114 tickets

PLZ MARK ME BRAINLIEST

4 0
3 years ago
Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a
omeli [17]

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287

This probability is 1 subtracted by the pvalue of Z when X = 900.

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

By the Central Limit Theorem

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

Z = \frac{900 - 880}{9.1287}

Z = 2.19

Z = 2.19 has a pvalue of 0.9857

1 - 0.9857 = 0.0143

1.43% probability that the sample mean rent price will be greater than $900

8 0
3 years ago
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