Best explained and correct answer gets brainliest!

Need help ASAP !!!!!!

Slav-nsk [51]

Answer:

Therefore, Point M( 1 , -2 ) is the Mid point of segment ST.

Step-by-step explanation:

Given:

Let,

point S( x₁ , y₁) ≡ ( -1 , 1)

point T( x₂ , y₂) ≡ (3 , -5)

Point M( x , y ) is the Mid point of segment ST.

To Find:

Point M( x , y )= ?

Solution:

As Point M( x , y ) is the Mid point of segment ST.

So we have Mid Point Formula as

$Mid\ pointM(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

On substituting the given values in above equation we get

$M(x,y)=(\frac {-1+3}{2},\frac{-5+1}{2})\\\\M(x,y)=(\frac{2}{2},\frac{-4}{2}) \\\\M(x,y)=(1,-2)\ \textrm{which the required midpoint}$

Therefore, Point M( 1 , -2 ) is the Mid point of segment ST.

4 0

3 years ago

How many zeros are there in the standard form of six hundred thousand, twenty?

Julli [10]

There are Three zeros

3 0

3 years ago

11) A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randoml

Illusion [34]

Answer:

The confidence interval would be more narrow if the confidence level were changed to 90%

Step-by-step explanation:

For a small sample size of n = 18 a pivotal quantity that we can use to form a confidence interval for $\mu$ is given by $T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$ that has a t distribution with (n-1) degrees of freedom. We find a $100(1-\alpha)$ % confidence inverval using $P(-t_{\alpha/2}\leq T \leq t_{\alpha/2}) = 1-\alpha$ , where $t_{\alpha/2}$ is the t-value such that there is an area equal to $\alpha/2$ above this t-value and below the curve of the density of the t distribution with n-1 df. We find a 95% confidence interval with $P(-t_{0.025}\leq T\leq t_{0.025}) = 0.95$ and we find a 90% confidence interval with $P(-t_{0.05}\leq T\leq t_{0.05}) = 0.90$ . Because of $-t_{0.025} < -t_{0.05}$ and $t_{0.05} < t_{0.025}$ , the confidence interval would be more narrow if the confidence level were changed to 90%.