Answer: <span><span>2x² + x - 2</span> (the first option)
Explanation:
1) Question: divide 8x⁴+2x³-7x²+3x-2 by 4x² - x + 1
2) First term of the quotient
</span><span> 8x⁴ + 2x³ - 7x² + 3x - 2 | 4x² - x + 1
---------------------
-8x⁴ + 2x³ - 2x² 2x²
----------------------------------
4x³ - 9x² + 3x - 2
3) Second term of the quotient:
</span>
<span> 8x⁴ + 2x³ - 7x² + 3x - 2 | 4x² - x + 1
---------------------
-8x⁴ + 2x³ - 2x² 2x² + x
----------------------------------
4x³ - 9x² + 3x - 2
-4x³ + x² - x
----------------------------
- 8x² + 2x - 2
4) third term of the quotient:
</span>
<span> 8x⁴ + 2x³ - 7x² + 3x - 2 | 4x² - x + 1
---------------------
-8x⁴ + 2x³ - 2x² 2x² + x - 2
----------------------------------
4x³ - 9x² + 3x - 2
-4x³ + x² - x
----------------------------
- 8x² + 2x - 2
8x² - 2x + 2
-------------------------
0
5) Conclusion: since the remainder is 0, the division is exact and the quotient is </span>2x² + x - 2
You can verify the answer by multiplying the quotient obtained by the divisor. The result has to be the dividend.
Answer:
C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.
Step-by-step explanation:
Let μ be the the average attendance at games of the football team
The claim: the average attendance at games is over 523
Null and alternative hypotheses are:
: μ=523
: μ>523
The conclusion is failure to reject the null hypothesis.
This means that <em>test statistic</em> is lower than <em>critical value</em>. Therefore it is not significant, there is no significant evidence to accept the <em>alternative</em> hypothesis.
That is no significant evidence that the average attendance at games of the football team is greater than 523.
Answer:
not sure if it's ryt but I think it's A
Using pseudocode:
printArray(arr[], integers)
DECLARE integers
integers = SizeOf(arr)
FOR i = 1 to integers // loop from 1 to the number of elements in arr[]
print(i)
print('')
i = i + 1
ENDFOR
END
Answer:
The answer is D: the number of candies in a box.
Step-by-step explanation:
