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MA_775_DIABLO [31]
3 years ago
6

Write four numbers that round to 700,000 when rounded to the nearest hundred thousand

Mathematics
1 answer:
slega [8]3 years ago
6 0
699,995
699,996
699,997
699,998
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Simplest form of the expression -2x2(x – 5) + x(2x2 – 10x) + x is
Andrei [34K]
You first simplify the expression using PEMDAS
-2x^3-10x^2+2x^3-10x^2+x
then combine like terms,
(-2x^3+2x^3), (-10x^2-10x^2), x
cancels out^
so the answer would be -20x^2+x
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3 years ago
A certain bacteria population increases continuously at a rate proportional to its current number. The initial population of the
luda_lava [24]

Answer: C

Step-by-step explanation:

This describes an expotential function

At t = 0, P = 70

Therefore, C = 70

P = 70 e^kt

Solve for k by plugging in (4,360)

k = 0.4094

plug in t(7 hours)

70 e^(0.4094*7)

The answer roughly equals C

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3 years ago
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Factor Completely: x2 - 81 <br> A.(x-9)(x+9)<br> B.(x+9)(x+9)<br> C.(x-9)(x-9)<br> D.(9-x)(9+x)
Hunter-Best [27]

Answer is A

The binomial can be factored using the difference of squares formula a2-b2 = (a+b) (a-b) where a=x and b=9

(X+9) x(x-9)

7 0
2 years ago
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onsider the following hypothesis test: H 0: 50 H a: &gt; 50 A sample of 50 is used and the population standard deviation is 6. U
kondaur [170]

Answer:

a) z(e)  >  z(c)   2.94 > 1.64  we are in the rejection zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c)  1.18 < 1.64  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12  > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point  z(c) ;  α = 0.05  fom z table w get   z(c) = 1.64 we need to compare values (between z(c)  and z(e) )

The test hypothesis is:  

a) H₀      ⇒      μ₀  = 50     a)  Hₐ    μ > 50   ;    for value 52.5

                                          b) Hₐ    μ > 50   ;     for value 51

                                          c) Hₐ    μ > 50   ;      for value 51.8

With value 52.5

The test statistic    z(e)  ??

a)  z(e) =  ( μ  -  μ₀ ) /( σ/√50)      z(e) = (2.5*√50 )/6   z(e) = 2.94

2.94 > 1.64  we are in the rejected zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) =  ( μ  -  μ₀ ) /( σ/√50)    ⇒  z(e) =  √50/6    ⇒  z(e) = 1.18

z(e) < z(c)  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e)  =  ( μ  -  μ₀ ) /( σ/√50)    ⇒ z(e)  = (1.8*√50)/ 6   ⇒ z(e) = 2.12

2.12  > 1.64 and we can conclude the same as in case a

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–3x + 1 + 10x = x + 4
Gnom [1K]

Answer:

Exact fraction form: 1/2 Decimal form: 0.5

8 0
2 years ago
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