7. y = 2x² + 3 function not function ?

Can someone please help me with this plz only on 13

BaLLatris [955]

Answer:

2/9

Step-by-step explanation:

To determine the fraction of students that walk to school, take the fraction of students that do not ride the bus and multiply by the fraction that walk

6/15* 5/9

Simplify the fractions

6/15 Divide the top and bottom by 3

2/5

2/5 * 5/9 = 2/9

4 0

3 years ago

Which of these tables represents a non-linear function?

olchik [2.2K]

Answer:

we conclude that the table ''A 2-column table with 4 rows. Column 1 is labeled x with entries 17, 18, 19, 20. Column 2 is labeled y with entries 16, 17, 19, 20'' represents a non-linear function.

Step-by-step explanation:

The nature of whether a function can be linear or non-linear, we must check how the x and y values of the table change. If the first difference between y-values remains the same (constant first difference), then the table will represent a linear function, otherwise not.

Given the x and y entries of the first table:

x y

17 20

18 19

19 18

20 17

From the table, it is clear that as x constantly increases by 1 unit, the y-values are also changing constantly by 1 unit. The first difference between y-values remains the same.

i.e. 19-20=-1, 18-19=-1, 17-18=-1

Thus, this table represents a linear function.

Given the x and y entries of the second table:

x y

17 -16

18 -17

19 -18

20 -19

From the table, it is clear that as x constantly increases by 1 unit, the y-values are also changing constantly by 1 unit. The first difference between y-values remains the same.

i.e. -17-(-16)=-1, -18-(-17)=-1, -19-(-18)=-1

Thus, this table represents a linear function.

Given the x and y entries of the second table:

x y

17 16

18 17

19 19

20 20

From the table, it is clear that as x constantly increases by 1 unit, but the y-values are not changing constantly. The first difference between y-values does not remain the same.

i.e. 17- 16 = 1, 19 - 17 = 2, 20 - 19 = 1

Thus, this table does not represent a linear function. Hence, it is a non-linear function.

Given the x and y entries of the second table:

x y

17 -20

18 -19

19 -18

20 -17

From the table, it is clear that as x constantly increases by 1 unit, the y-values are also changing constantly by 1 unit. The first difference between y-values remains the same.

i.e. -19-(-20)=1, -18-(-19)=1, -17-(-18)=1

Thus, this table represents a linear function.

Therefore, we conclude that the table ''A 2-column table with 4 rows. Column 1 is labeled x with entries 17, 18, 19, 20. Column 2 is labeled y with entries 16, 17, 19, 20'' represents a non-linear function.

4 0

3 years ago

The point M(-6, -4) is translated 2 units right. What are the coordinates of the resulting point, M′?

fredd [130]

B) (-4, -4)

When moving to the right you are moving to a more POSITIVE region on the x-axis; meaning, the y-value (vertical axis) does not get affected unless you’re moving up or down.

Remember: (-6, -4)
X = -6
Y = -4

If you move 2 units to the right (x-axis) you go from -6 + 2 which equals -4. And again, you’re not moving up and down so your
y-value stays the same and your new coordinates are (-4, -4)

7 0

3 years ago

Mai bought a soccer ball and 2 pairs of shorts. how much will she pay for everything including tax? what information will you ne

Andre45 [30]

Answer:

You would need to find out the cost of the soccer ball and the price for shorts. Then you will need to find out how much tax she will pay.

Step-by-step explanation:

5 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

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 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.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

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

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

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

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

3 years ago