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Firdavs [7]
3 years ago
14

In circle O, central angle AOB has a measure of 90°. Which of the following is not true? segment OA is a radius segments OA and

OB are perpendicular arc AB is a semicircle

Mathematics
2 answers:
Zinaida [17]3 years ago
5 0

Answer:

Arc AB is a semicircle, Not true.

Step-by-step explanation:

In circle O, central angle AOB has measure of 90°

We need to choose statement which is not true. We can see the attachment to shape of circle.  

As we can see O is the center of circle. A and B are two points on circle.  

So, OA and OB must be radius of this circle.  

1) Segment OA is a radius ( TRUE )

2) Segment OA and OB are perpendicular (TRUE)

Because central angle makes 90°  

3) arc AB is a semicircle (FALSE)

Because AOB is not 180°


lesantik [10]3 years ago
3 0
The last one: the arc AB is a semicircle. 

a semicircle is 180 degrees
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Grace had $51 when she browsed an on-line store, which sells a pair of green earrings for $9.00 each plus a one-time $6.00 shipp
vitfil [10]

Answer:

5 Green earrings

Step-by-step explanation:

Given that:

Total amount held = $51

Cost per pair of green earrings = $9

One time shipping fee = $6

How many pairs of green earrings did grace purchase?

Let the number of green earrings purchased = g

Then,

$9*g + $6 = $51

$9g + $6 = $51

$9g = $51 - $6

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g = $45 / $9

g = 5

Hence, the number of green earrings purchased = 5

4 0
3 years ago
Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o
kirza4 [7]

Answer:

a) H0: \sigma = 1.34

H1: \sigma \neq 1.34

b) df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

c) t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0
2 years ago
What is the answer to the math problem 5+9(2+4)?
IceJOKER [234]

Answer:

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5 + 54

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