What values of a make the equation have non real solution?

Find the rate of change between (-3,6) (0,-3)

marin [14]

<h3>The rate of change between (-3,6) (0,-3) is -3</h3>

Solution:

Given that,

We have to find the rate of change between (-3,6) (0,-3)

The rate of change is given as:

$Rate\ of\ change = \frac{y_2-y_1}{x_2-x_1}$

From given,

$(x_1, y_1) = (-3, 6)\\\\(x_2, y_2) = (0, -3)$

Substituting the values we get,

$Rate\ of\ change = \frac{-3-6}{0+3}\\\\Rate\ of\ change = \frac{-9}{3}\\\\Rate\ of\ change = -3$

Thus rate of change is -3

3 0

3 years ago

PLEASE HELP 20 POINTS!!!!!!!

svetoff [14.1K]

Answer:

18/25

Step-by-step explanation:

i dont have an explanation

3 0

3 years ago

Please help with this question please

puteri [66]

Answer:

Choice A

Step-by-step explanation:

(4m^5n^2/m^2n)^3

dividing exponents = subtraction

(4m^3n)^3

4^3 = 64

(m^3 )^3 = m^3x3 = m^9

64m^9n^3

5 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

Enter the equation of the line in slope-intercept form.

Karo-lina-s [1.5K]

Answer:

y = 1/2x + 1

Step-by-step explanation:

Slope: 1/2

Point: (2,2)

y-intercept: 2 - (1/2)(2) = 2 - 1 = 1

4 0

2 years ago