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svp [43]
2 years ago
7

8

Mathematics
2 answers:
Usimov [2.4K]2 years ago
6 0

Answer:

This will help you- onlinemath4all.com/how-to-find-the-distance-of-a-chord-from-the-center-of-a-circle.html

If u still can't solve it msg me in this answer.

jenyasd209 [6]2 years ago
5 0

answer is 5 cm

I am happy to help

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Pls help asap
yuradex [85]

Answer:

y=525x+230

Step-by-step explanation:

525 is spent everyday. x is the number of days, so with each day $525 is spent.

$230 is a one time cost, regardless of how many days they stay on the trip.

3 0
3 years ago
Fraction 4 over 5 n = Fraction 2 over 3 .n = ___?
xxTIMURxx [149]
The answer is b. Hope this was helpful
3 0
2 years ago
Read 2 more answers
If you borrow $100 for 3 years with an annual interest rate of 9% how much will you pay altogether.
arsen [322]
100(1.09)^3=129.5029 or $129.50

3 0
3 years ago
Read 2 more answers
The condition_______?proves that ∆ABC and ∆EFG are congruent by the SAS criterion.
snow_lady [41]

Answer:

(1)  D.Angle C is congruent to to Angle F. (2) C. SSS. (3) C. cannot be congruent to.

Step-by-step explanation:

1)

From the given figure it is noticed that

AC=EG

CB=GF

According to SAS postulate,  if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then both triangles are congruent.

The included angles of congruent sides are angle C and angle G.

So, condition "Angle C is congruent to to Angle F"  will prove that the ∆ABC and ∆EFG are congruent by the SAS criterion.

2)

If AB\neq EF

According to SSS postulate,  if all three sides in one triangle are congruent to the corresponding sides in the other.

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore SSS criterion for congruence is violated.

3)

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore the included angle of congruent sides are different.

\angle C\neq \angle G

Therefore angle C and angle F cannot be congruent to each other.

4 0
3 years ago
A book claims that more hockey players are born in January through March than in October through December. The following data sh
denis-greek [22]

Complete question :

Birth Month Frequency

January-March 67

April-June 56

July-September 30

October-December 37

Answer:

Yes, There is significant evidence to conclude that hockey​ players' birthdates are not uniformly distributed throughout the​ year.

Step-by-step explanation:

Observed value, O

Mean value, E

The test statistic :

χ² = (O - E)² / E

E = Σx / n = (67+56+30+37)/4 = 47.5

χ² = ((67-47.5)^2 /47.5) + ((56-47.5)^2 /47.5) + ((30-47.5)^2/47.5) + ((37-47.5)^2/47.5) = 18.295

Degree of freedom = (Number of categories - 1) = 4 - 1 = 3

Using the Pvalue from Chisquare calculator :

χ² (18.295 ; df = 3) = 0.00038

Since the obtained Pvalue is so small ;

P < α ; We reject H0 and conclude that there is significant evidence to suggest that hockey​ players' birthdates are not uniformly distributed throughout the​ year.

5 0
2 years ago
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