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Alexxandr [17]
2 years ago
12

(Please help )Prove or disprove that point (4.00, 4,50) lies on circle O centered at the origin.

Mathematics
1 answer:
Marat540 [252]2 years ago
4 0

Answer:

go ahead and view TaraLM she's making free credit reports on her phone she's on the contact list this is

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Find (f°g)(x) f(x)=x^2-1, g(x)=3x-2
Bess [88]
F°g(x) = (3x-2)² -1
= 3x(3x-2)-2(3x-2) -1
= 9x² -6x-6x+2-1
=9x²-12x+1
5 0
3 years ago
This year’s school population in Waterloo is 135 percent of last year’s school population. This year’s student population is 756
sp2606 [1]

Answer:560

Step-by-step explanation:

Let a be the number of students the school have last year

135% of a=756

135/100 x a=756

135a/100=756

Cross multiplying we get

135a=756 x100

135a=75600

Divide both sides by 135

135a/135=75600/135

a=560

6 0
3 years ago
A social psychologist predicted that ratings of an individual’s social desirability would be influenced by their physical attrac
marshall27 [118]

Answer:

It is stated in the query that the more attractive the individual, the higher their rank of desirability.

Step-by-step explanation:

This means that we should conclude that the alternative explanation of "higher degree of desire contributes to higher levels of social attractiveness"

In the event of a null hypothesis, we will believe that the degree of attractiveness has no impact on the score of cultural desirability.

(1) Null hypothesis:- The level of attraction does not have an impact on the rating of social desirability.

(2) Alternate hypothesis:- The level of attraction has a positive effect on the ranking of social desirability.

(3) The error of Type I is characterized as the rejection of the true null hypothesis. Therefore, in this case, the type I mistake would be that "the level of attraction has a positive effect on the rating of social desirability while the level of attraction does not have an impact on the rating of social desirability."

(4) The error of type II is characterized as the failure to reject the false null hypothesis. Thus, in this situation, the Type II error would be that "the level of attraction has no impact on the rating of social desirability while the real level of attraction has a positive effect on the rating of social desirability."

(5) We know that the likelihood of having a Form I error is equal to the degree of significance (in percent terms). The significance level, in this case, is 0.05, so the likelihood of making a form I error is 0.05 * 100=5 percent.

Therefore, there is a 5 percent risk of making a Type I mistake.

3 0
3 years ago
(r-3)(r-1)<br> Help me please!!!
madreJ [45]

Answer:

= r²−4r+3

Step-by-step explanation:

(r - 3) (r - 1)

(r x r) + (r x -1) + (-3 x r) + (-3 x -1)

r² + - r - 3r + 3

= r²−4r+3

5 0
3 years ago
](6^3 - 9)/23]4<br><br> Evaluate this expression
Andrej [43]
His iso youno Hepworth 6 sih
8 0
3 years ago
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