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notka56 [123]
3 years ago
10

What is the solution to the inequality |2n+5|>1​

Mathematics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

ANSWER

n < - 3 \: or \: n > - 2n<−3orn>−2

EXPLANATION

The given inequality is,

|2n + 5| \: > \: 1∣2n+5∣>1

By the definition of absolute value,

- (2n + 5) \: > \: 1 \: or \: (2n + 5) \: > \: 1−(2n+5)>1or(2n+5)>1

We divide through by negative 1, in the first part of the inequality and reverse the sign to get,

2n + 5 \: < \: - 1 \: or \: (2n + 5) \: > \: 12n+5<−1or(2n+5)>1

We simplify now to get,

2n \: < \: - 1 - 5 \: or \: 2n \: > \: 1 - 52n<−1−5or2n>1−5

2n \: < \: - 6 \: or \: 2n \: > \: - 42n<−6or2n>−4

Divide through by 2 to obtain,

n \: < \: - 3 \: or \: n \: > \: - 2n<−3orn>−2

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3.4 grams

Step-by-step explanation:

The water 20g has salt solution of 17%.

If you were to evaporate all the water.

17% × 20

0.17 × 20 = 3.4

You would get 3.4 grams of salt.

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If f(x) = 4x – 3 and g(x) = x + 4, find (f – g)(x).
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3 years ago
A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0
3 years ago
Someone please help! find x round to the nearest tenth! trigonometry!​
TEA [102]

Answer:

<h2>x = 745.5 ft</h2>

Step-by-step explanation:

Look at the picture.

Use sine:

sine=\dfrac{opposite}{hypotenuse}

We have a measure of the angle and length of the opposite:

\alpha=28^o,\ opposite=350\ ft

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Substitute:

\dfrac{350}{x}=0.4695

\dfrac{350}{x}=\dfrac{4695}{10000}            <em>cross multiply</em>

4695x=3500000           <em>divide both sides by 4695</em>

x=\dfrac{3500000}{4695}\to x\approx745.5

5 0
3 years ago
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