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Sedbober [7]
3 years ago
15

use the vertical line test to determine if the relation (-6,-2), (-2,6), (0,3), (3,5) is a function. Explain your response.

Mathematics
1 answer:
sattari [20]3 years ago
6 0
Graph the points and connect them and draw vertical lines in between them. You can already tell it's a function because every input has exactly on output.
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Does anybody know the answer?
kirill [66]

Answer:

y+48=90

y= 90-48

y= 42

Step-by-step explanation:

ok

by using the property

sum of two opposite interior angles of triangle is equal to exterior angle

4 0
2 years ago
If y = -12 when x = 9, find u when x = -4
True [87]

Answer:

y should equal -25 if i did the math right

Step-by-step explanation:

-12 and 9 have a difference of -21 so if i subtract -21 from -4 it would equal -25.. hope this helps!!

4 0
2 years ago
A motor vehicle has a maximum efficiency of 39 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph
svp [43]

Answer:

38 mpg

Step-by-step explanation:

Initial efficiency = 39 mpg

Initial speed = 40 mph

Final speed = 50 mph

The efficiency drop  between 40 mph and 50 mph is given by:

E_d = 0.1 \frac{mpg}{mph}*\Delta V

The total efficiency drop from 40 to 50 mph is:

E_d = 0.1 \frac{mpg}{mph}*(50-40)mph\\E_d = 1\ mpg

Therefore, the efficiency at 50 mph is:

E_{50} = E_{40} -E_d\\E_{50} = 39 -1\\E_{50} = 38\ mpg

5 0
3 years ago
I NEED HELP PLEASE EXPLAIN
kari74 [83]

Answer:

The function concerning the evolution of the value of the first car is exponential and the one concerrning the value of the second car is linear

Step-by-step explanation:

1st car's drop in value ( not considering the second hand type of drop):

1.4500$

2.4050$

3.3645$

2nd car's value is dropping 3000$ every year so the function is f(n)=45000-3000n, n- number of years

6 0
2 years ago
Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 67.7 grams an
Alexxx [7]

Answer:

<u>a. s.e. = 0.338</u>

<u>b. The  probability of finding a random slice of pizza with a mass of less than 67.2 grams is 0.5871 or 58.71%</u>

<u>c. The probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams is 0.0694 or 6.94%</u>

d. <u>The sample mean of 62.75 would represent the 15th percentile</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question properly:

μ of the mass of slices of pizza for a certain brand = 67.7 grams

σ of the mass of slices of pizza for a certain brand = 2.28 grams

2. For samples of size 20 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?

Let's recall that the standard deviation of the sampling distribution of the mean is called the standard error of the mean and its formula is:

μσs.e.= √σ/n, where n is the sample size.

s.e. = √2.28/20

<u>s.e. = 0.338</u>

3. What is the probability of finding a random slice of pizza with a mass of less than 67.2 grams?

Let's find the z-score for X = 67.2, this way:

z-score = (X - μ)/σ

z-score = (67.2 - 67.7)/2.28

z-score = 0.5/2.28

z-score = 0.22 (rounding to the next hundredth)

Now, using the z-table, let's find p, the probability:

p (z = 0.22) = 0.5871

<u>The  probability of finding a random slice of pizza with a mass of less than 67.2 grams is 58.71%</u>

4.  What is the probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams?

Let's use the central limit theorem to find the z-score, this way:

z-score = X - μ/s.e

z-score = 67.2 - 67.7/0.338

z-score = -0.5/0.338

z-score = - 1.48

Now, using the z-table, let's find p, the probability:

<u>p (z = -1.48) = 0.0694</u>

5. What sample mean (for a sample of size 20) would represent the bottom 15% (the 15th percentile)?

For p = 0.150, let's find the z-score:

z-score = - 2.17

z-score = X - μ/s.e

- 2.17 = (X - 67.7)/2.28

- 4.95 = X - 67.7

X = 67.7 - 4.95

X = 62.75 (rounding to the next hundredth)

<u>The sample mean of 62.75 would represent the 15th percentile</u>

7 0
2 years ago
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