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2 years ago

This pattern can be represented by the explicit expression of 2n.

Mathematics

2 answers:

max2010maxim [7]2 years ago

8 0

Answer:

its is false

Step-by-step explanation:

the sequence would be 0,2,4,6

vesna_86 [32]2 years ago

6 0

Answer:

False

Step-by-step explanation:

The given sequence for n = 0, 1, 2, 3 is 0, 1, 4, 9. The suggested expression for the pattern is 2n. If we use the given values of n in that expression, we get a sequence of 0, 2, 4, 6. That is not the same sequence.

The explicit expression does not represent the given pattern. (False)

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It's 1/2,-2 just did the math.

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3 years ago

Read 2 more answers

What is the area of this figure?

timurjin [86]

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Step-by-step explanation:

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3 years ago

Items produced by a manufacturing process are supposed to weigh 90 grams. The manufacturing procon

Annette [7]

Using the normal distribution, it is found that 0.26% of the items will either weigh less than 87 grams or more than 93 grams.

In a <em>normal distribution</em> with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of 90 grams, hence $\mu = 90$ .

The standard deviation is of 1 gram, hence $\sigma = 1$ .

We want to find the probability of an item <u>differing more than 3 grams from the mean</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3}{1}$

$Z = 3$

The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.

Looking at the z-table, Z = -3 has a p-value of 0.0013.

2 x 0.0013 = 0.0026

0.0026 x 100% = 0.26%

0.26% of the items will either weigh less than 87 grams or more than 93 grams.

For more on the normal distribution, you can check brainly.com/question/24663213

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3 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

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3 years ago