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Alex_Xolod [135]
3 years ago
11

Factorise 12x^3-4x^2

Mathematics
1 answer:
Kobotan [32]3 years ago
5 0
You factor out the number of common x's and the common number 4.

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Luke earned $36,000 during the first year of his job at Putt-Putt Kingdom. After each year he received a 10% raise. Find his tot
Lady_Fox [76]
In order to solve this problem, you need to use a geometric series:
S_{n} =  \frac{ a_{1}(1 - r^{n}) }{1 - r}
where:
a₁ = first term of the series = 36000
r = common rate = 10% raise, therefore 1.10
n = number of terms = 5

Therefore,
 <span>S_{n} =  \frac{ 36000(1 - 1.10^{5}) }{1 - 1.10}
= 219783.60 $

Luke's total earnings in five years are <span>219783.60 $.</span>

</span>
6 0
3 years ago
1/3 of 90 = 2/3 of ?
rewona [7]
Answer=45
1/3 of 90= 30
5 0
3 years ago
PLZ HELP ASAP SOLID GEOMETRY
Vera_Pavlovna [14]
Volume of the cylinder can be calculated as:

V=πr²h

Radius = r = 4
Height = h = 7
π = 3.14

Using the values, we get:

V = 3.14 (16) (7) = 351.68 cubic units

So the answer to this question is 351.68
7 0
3 years ago
Read 2 more answers
The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control a
Dafna11 [192]

Answer:

Probability that at least 490 do not result in birth defects = 0.1076

Step-by-step explanation:

Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) = \sum\limits^{500}_{x=490} {^{500} C_{x} } (\frac{32}{33} )^{x} (\frac{1}{33} )^{500-x}

                 = {^{500} C_{490} } (\frac{32}{33} )^{490} (\frac{1}{33} )^{500-490}  + .......+ {^{500} C_{500} } (\frac{32}{33} )^{500} (\frac{1}{33} )^{500-500}

                = 0.04541 + ......+0.0000002079

                = 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076

4 0
2 years ago
How do i factor <br><br> a2-b2
bagirrra123 [75]
(a^2-b^2) = (a+b)(a-b)

You can use this equation for any numbers or variables in the form a2 - b2.
6 0
3 years ago
Read 2 more answers
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