Please help!!!

What is the distance between the points (-4,2) and (3,-5)

34kurt

Answer:

7√2 units

Step-by-step explanation:

The distance between two points having coordinates (x1, y1) and (x2, y2) is given by,

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

Here, x1=-4 and y1=2

x2=3 and y2=-5

$\sqrt{ {(3 - ( - 4))}^{2} + {( - 5 - 2)}^{2} }$

$\sqrt{ {(3 + 4)}^{2} + {( - 7)}^{2} }$

$\sqrt{ {(7)}^{2} + {( - 7)}^{2} }$

$\sqrt{49 + 49} = 7 \sqrt{2}$

6 0

4 years ago

Read 2 more answers

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

The vertical asymptote of the function f(x) = 3 log(x + 3)

ipn [44]

A function has a vertical asymptote where it is not defined, and the logarithm is defined only if its argument is strictly positive: it has a vertical asymptote if its argument is zero, and is not defined if its argument is negative.

So, you have a vertical asymptote where

$x+3=0 \iff x=-3$