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harina [27]
3 years ago
5

Is 0.024 and .024 the same thing

Mathematics
2 answers:
serious [3.7K]3 years ago
7 0
Yes they are the same thing. 
DerKrebs [107]3 years ago
6 0
No it is not the same thing

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The table shows the low outside temperatures for Monday, Tuesday, and Wednesday.
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-8.8 - (16.5) =
-8.8 + 16.5 =
7.7 <==
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4 years ago
A farmer picked 1,060 blueberries equally from 5 rows of plants. How many blueberries did he pick from each row?
Nataly_w [17]
1,060 ÷ 5 = 212. He picked 212 blueberries from each row
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Nikhil asked 120 randomly chosen moviegoers to watch a clip of an upcoming movie and choose the best title. The title "Everythin
aleksandrvk [35]
The mean proportion is p = 36/120 = 0.3.
The standard deviation of the proportion will be sqrt(p*(1-p)/n) = sqrt(0.3*0.7/120) = 0.0418. We multiply this by the z-score of 1.645 to get a deviation of 0.0688.
Therefore, the confidence interval is (0.3 - 0.0688, 0.3 + 0.0688), which is
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8 0
3 years ago
Read 2 more answers
Can any one help me 11 points no lnks pls. Just help with any of these
zvonat [6]

Answer:

5. m<1 = 114 degrees

   m<2 = 66 degrees

6. 2x + 3x + 4x = 9x = 180, x = 20

7. 3

8. 1/3

<em>Hope that helps!</em>

<em>-scsb</em>

Step-by-step explanation:

8 0
3 years ago
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For the polynomial function ƒ(x) = −x6 + 3x4 + 4x2, find the zeros. Then determine the multiplicity at each zero and state wheth
murzikaleks [220]

There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.


You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.


-x^6 + 3x^4 + 4x^2

-x^2(x^4 - 3x^2 - 4)


Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.


-x^2(x^4 - 3x^2 - 4)

-x^2(x^2 - 4)(x^2 + 1)


Now we can use the factors of two perfect squares rule to factor the middle parenthesis.


-x^2(x^2 - 4)(x^2 + 1)

-x^2(x - 2)(x + 2)(x^2 + 1)


We would also want to split the term in the front.


-x^2(x - 2)(x + 2)(x^2 + 1)

(x)(-x)(x - 2)(x + 2)(x^2 + 1)


Now we would set each portion equal to 0 and solve.


First root

x = 0 ---> no work needed


Second root

-x = 0 ---> divide by -1

x = 0


Third root

x - 2 = 0

x = 2


Forth root

x + 2 = 0

x = -2


Fifth and Sixth roots

x^2 + 1 = 0

x^2 = -1

x = +/- \sqrt{-1}

x = +/- i

7 0
3 years ago
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