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Aleksandr [31]
2 years ago
6

There were 4 car races on saturdayand 3 on Sunday. If there were 9 cars racing in each race, how many cars raced over the two da

ys?
Mathematics
1 answer:
sergij07 [2.7K]2 years ago
3 0
There were 63 cars in each race.

You do the distributive property which would look like 9(3+4). Once you distribute that you should have 27+36. You add that and get the answer of 63. Hope this helps!
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Paul exercises for h hours a day for five days. Over the weekend, he exercises a total of 4.5 hours.
kirill [66]

Answer: 5h+4.5 i got it right on edg

Step-by-step explanation:

6 0
2 years ago
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Your sample size is 10 and your confidence is 90%. What is your t-score?
PilotLPTM [1.2K]

Answer:

if you want a t*-value for a 90% confidence interval when you have 9 degrees of freedom, go to the bottom of the table, find the column for 90%, and intersect it with the row for df = 9. This gives you a t*–value of 1.833 (rounded).

Step-by-step explanation:

Hope this helps!

6 0
3 years ago
ANSWER ASAP PLEASE !!!
Sauron [17]
3/7 = 0.43 (approx.)
4/9 = 0.44 (approx.)

Hence, 4/9 is bigger. 
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3 years ago
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What number must be added to complete the square? X^2-20x=17
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4 0
3 years ago
Factored 8x2y3-5= <br> I need the answer
ANTONII [103]

Answer: Reformatting the input :

Changes made to your input should not affect the solution:

(1): "y3" was replaced by "y^3". 1 more similar replacement(s).

STEP

1

:

Equation at the end of step 1

(23x2 • y3) - 5

STEP

2

:

Trying to factor as a Difference of Cubes

2.1 Factoring: 8x2y3-5

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : 8 is the cube of 2

Check : 5 is not a cube !!

Ruling : Binomial can not be factored as the difference of two perfect cubes

Final result :

8x2y3 - 5

5 0
3 years ago
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