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3 years ago

What is 28.204 rounded to the nearest whole

Mathematics

2 answers:

SOVA2 [1]3 years ago

8 0

Answer:

Step-by-step explanation:

Ray Of Light [21]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is the range of values of x for a triangle's third side given side measures of 8 and 15 for the other two sides?

mr_godi [17]

Answer: b

Step-by-step explanation:

it is right

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3 years ago

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5. Find the thickness of each layer. Use the data to determine the percentage of the total

zmey [24]

Answer:

Exosphere - 10,000 kilometers thick

Thermosphere - 513 kilometers

Mesosphere - 2200 km

Stratosphere - Depends, but most occurring thickness is 50km

Troposphere - 8-14km

Step-by-step explanation:

Double check and verify this information with your lesson since honestly the numbers could very well differ depending on your lesson and when they were taken. I hope this helps though..have a good one

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2 years ago

What is the factored form of this expression?<br><br> 18x2 − 50

BARSIC [14]

Answer:

_14

Step-by-step explanation:

step1 18×2 -50

step2 18×2 =36_50

step3 36_50=_14

8 0

3 years ago

Read 2 more answers

From Euler’s relation eiθ = cosθ + isinθ,

adelina 88 [10]

Answer:

(a) The graphic representation is in the attached figure.

(b) $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ .

(c) $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ .

Step-by-step explanation:

(a) Given a complex number $e^{i\theta}$ we know, from Euler's formula that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ . So, it is not difficult to notice that

$|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1$

so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is $(\cos(\theta), \sin(\theta))$ .

Now,

$e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)$ .

Notice that $e^{-i\theta}$ has the same modulus that $e^{i\theta}$ , so it is on the unit circumference. Beside, its Cartesian representation is $(\cos(\theta), -\sin(\theta))$ .

So, the points $(\cos(\theta), \sin(\theta))$ and $(\cos(\theta), -\sin(\theta))$ are symmetric with respect to the X-axis. All this can be checked in the attached figure.

(b) Notice that

$e^{i\theta} + e^{-i\theta} = \cos(\theta)+i\sin(\theta) + \cos(\theta)-i\sin(\theta) = 2\cos(\theta)$

Then,

$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ .

(c) Notice that

$e^{i\theta} - e^{-i\theta} = \cos(\theta)+i\sin(\theta) - \cos(\theta)+i\sin(\theta) = 2i\sin(\theta)$

Then,

$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ .

7 0

3 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 120 customer orders to f

Aliun [14]

Answer:

a. P(X = 0) = 0.02586

b. $\mathbf{P(X \leq 2 ) =0.2879}$

c. $\mathbf{P(X \leq 5 ) =0.8387}$

Step-by-step explanation:

From the given information:

a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X = 0)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}$

$P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}$

P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰

P(X = 0) = 0.02586

b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ]$

$P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}]$ $P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}]$

$P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)]$

$\mathbf{P(X \leq 2 ) =0.2879}$

(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ]$

$P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}]$ $P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}]$

$\mathbf{P(X \leq 5 ) =0.8387}$

5 0

3 years ago