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3 years ago

9) Let X-2, x, x+3 be the first 3 terms of a geometric sequence. FInd the value of x.. PLEASE HELP

1 answer:

7 0

Hello,

Let's r the ration

We suppose x ≠0

x=r*(x-2)==>r=x/(x-2)

x+3=r*x==> (x+3)/x=x/(x-2)==>(x+3)(x-2)=x²
==>x²+3x-2x-6=x² ==>x-6=0==>x=6

x-2=6-4=4
x=6
x+3=6+3=9
6=4*3/2
9=6*3/2

Ration 3/2
First term: 4

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Hey Guys! I need some help. This is the question... Sara travels often for her job. The miles she traveled for the last seven wo

Schach [20]

Using the median concept, it is found that the interquartile range of Sara's daily miles is of 21 miles.

<h3>What are the median and the quartiles of a data-set?</h3>

The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.

The first quartile is the median of the first half of the data-set.

The third quartile is the median of the second half of the data-set.

The interquartile range is the difference of the quartiles.

The ordered data-set is given as follows:

65, 72, 86, 88, 91, 93, 97

There are 7 elements, hence the median is the 4th element, of 88. Then:

The first half is 65, 72, 86.

The second half is 91, 93, 97.

Since the quartiles are the medians of each half, the have that:

The first quartile is of 72 miles.

The third quartile is of 93 miles.

The interquartile range is of 93 - 72 = 21 miles.

More can be learned about the median of a data-set at brainly.com/question/3876456

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2 years ago

List all sets that apply to each number.

Tasya [4]

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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

4. Solve the equation for the given variable. For each step. Identity the property used to convert the equation. 22x + 11 = 4x -

Andrei [34K]

Answer:

x=-1

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

There are five cars in line at a stoplight. Each of these cars is a different color and different type of car. Diego is driving

adell [148]

Ok, here we go:
5 cars total.
1. Rachel - first car in line, yellow, SUV
2. blue car (Rachel is in front of the blue car), convertible, Anton
3. minivan - third car in line, Diego
4. white car, sedan, Ingrid
5. green car, immediately behind white car, Yuna, pickup

We know that Rachel is first, a blue car is second and the minivan is third. We know that 3, 4 cannot be a convertible since one is a minivan and the other is a sedan. We also know that a yellow car is in front of the convertible.

Based off of that we know that car 5 cannot be a convertible since 4 is white and not yellow. So 1 or 2 is the convertible and SUV. But we know that Rachel is in front of a blue car and the yellow car is in front of a convertible so that makes car #2 the blue convertible and car #1 the yellow SUV.

Fill in what you do know and start to eliminate the outliers. Hope that helps.

8 0

3 years ago