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3 years ago

Which of the following is NOT true of a perpendicular bisect or?

Mathematics

1 answer:

vova2212 [387]3 years ago

6 0

Answer:

The forth option

It forms a right angle with the segment.

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Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif

dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

Step:-1

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals $t_{0.01} =2.82$ (from tabulated value).

$The mean value = 0.0365$

$x±t_{0.01} \frac{S}{\sqrt{n-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{22} }$

$0.0365±2.82 \frac{0.0046}{4.690 }$

using calculator

$0.0365±0.00276$

Confidence interval is

$(0.0365-0.00276,0.0365+0.00276)$

$(0.0338,0.0392)$

the mean value is lies between in this confidence interval

(0.0338,0.0392).

Answer:-

using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.

4 0

2 years ago

What is the answer 5–4÷2?

tangare [24]

First, you divide 4 by 2 before you subtract them by 5 using BODMAS, or BIDMAS.

4 ÷ 2 = 2

5 - 2 = 3

3 is your answer

7 0

3 years ago

Read 2 more answers

Ill mark brainlyist!!!!

kumpel [21]

Answer: 30%

Step-by-step explanation:

First, we will put completed over the total.

$\frac{12}{40}$

Now, we will divide.

0.3

Lastly, we will multiply by 100 to turn it into a percent.

0.3 * 100 = 30%

You have ridden 30% of the course.

7 0

1 year ago

What is the midpoint for (-10,3) and (-3,-4)

vova2212 [387]

The midpoint is (-6.5,-0.5)

try using Desmos.com, when dealing with coordinates and functions. It really works!

7 0

2 years ago

Find the equation of the line that passes through (-2,3) and (4,-1).

horsena [70]

$Genera\ equation\ for\ line\\\\y=ax+b\\\\To\ find\ a\ and\ b\ substitude\ points\ (-2,3),\ (4,-1)\ into\ equation\\\\ \left \{ {{3=-2a+b}\ \ \ \ \atop {-1=4a+b\ |*-1}} \right.\\\\ \left \{ {{3=-2a+b}\ \ \ \ \atop {1=-4a-b\}} \right.\\+----\\\\Addition\ method\\\\4=-6a\ \ |:-6\\\\a=-\frac{4}{6}=-\frac{2}{3}\\\\b=3+2a=3+2* (-\frac{2}{3})=3-\frac{4}{3}=1\frac{2}{3}\\\\y=-\frac{2}{3}x+1\frac{2}{3}$

8 0

3 years ago