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gladu [14]
2 years ago
14

The ordered pairs model an exponential function, where w is the function name and t is the input variable. {(1, 60), (2, 240), (

3, 960), (4, 3840)}{(1, 60), (2, 240), (3, 960), (4, 3840)} What is the function equation in sequence notation? Enter your answer in the box.
Mathematics
2 answers:
kompoz [17]2 years ago
7 0
Hello there.
<span>
The ordered pairs model an exponential function, where w is the function name and t is the input variable. {(1, 60), (2, 240), (3, 960), (4, 3840)}{(1, 60), (2, 240), (3, 960), (4, 3840)} What is the function equation in sequence notation? Enter your answer in the box.
</span>
w = 15 (4^t)
alexandr1967 [171]2 years ago
3 0
We are given the set of points
<span>{(1, 60), (2, 240), (3, 960), (4, 3840)}{(1, 60), (2, 240), (3, 960), (4, 3840)} 
If w is the function name and t is the input variable, then the equation has the form
w = k a^t
Plugging in any two of the points to solve for k and a
Choosing </span>(1, 60) and (2, 240)<span>
60 = k a
k = 60 / a

240 = k a^2

Substituting
240 = 60/a (a^2)
a = 4
and
k = 60 / 4 = 15

The function is 
w = 15 (4^t)</span>
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The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci
Olenka [21]

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that n = 300 and \pi = \frac{10}{300} = 0.033

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532

The correct answer is

(0.0128, 0.0532)

4 0
1 year ago
Jones of Boston borrowed $40,000, on a 90-day 10% note. After 60 days, Jones made an initial payment of $6,000. On day 80, Jones
MaRussiya [10]

Answer:

$34,666.67

Step-by-step explanation:

Step 1: Calculate the interest for 60 days.

(40000*60/360 )* 0.10 = $666.67

Step 2: Amount = Principal + Interest = $40,000 + 666.67 = $40,666.67

Step 3 : After 60 days, he paid $6000.

Remaining balance of the first payment = 40666.67 - 6000 = $34,666.67

Thank you.

8 0
1 year ago
A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacemen
Elan Coil [88]

Answer:

  1. When Two containers are selected

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

    2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x 10^{-8} .  

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

                  Total containers = 445

                   Defective ones   = 3

           Non - Defective ones = 442 { Acceptable ones}

  • Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

  <em>Since we had selected one defective so for selecting second the available </em>

<em>   containers are 444 and available defective ones are 2 because once </em>

<em>    chosen they are not replaced.</em>

Hence, Probability that the second one selected is defective given that the first one was defective = \frac{2}{444} = 0.00450

(b) Probability that both are defective = P(first being defective) +

                                                                     P(Second being defective)

                 = \frac{3}{445} + \frac{2}{444} = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) +  P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

 So, Required Probability = \frac{442}{445}*\frac{441}{444} = 0.986

  • Three containers are selected, at random, without replacement from the batch.

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

<em>Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .</em>

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = \frac{1}{443} = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

<em>Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.</em>

Hence, Required probability = \frac{2}{443} = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

                              P(Second being defective) +  P(Third being defective)

        = \frac{3}{445}* \frac{2}{444}  * \frac{1}{443} = 6.855 x 10^{-8} .                

               

5 0
2 years ago
You by x packs of pencils, twice as many packs of erasers, and three times as many rolls of tape. Write an expression in the sim
ra1l [238]
Answer- since you are buying x packs of pencils, 2x packs of erasers, and 3x rolls of tape, you would be buying a total of 6x items.
7 0
1 year ago
The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. Find the probability
yawa3891 [41]

<u>Complete Question</u>

The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. The results are: 59 for Bride, 50 for Groom, 30 for both. Find the probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

Answer:

0.9875

Step-by-step explanation:

Total Number of Guests which forms the Sample Space, n(S)=80

Let the Event (a friend of the bride) =A

Let the Event (a friend of the groom) =B

n(A) =59

n(B)=50

Friends of both bride and groom, n(A \cap B)=30

Therefore:

n(A \cup B)=n(A)+n(B)-n(A \cap B)\\n(A \cup B)=59+50-30\\n(A \cup B)=79

The number of Guests who was a friend of the bride OR of the groom = 79

Therefore:

The probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

P(A \cup B) =\dfrac{n(A \cup B) }{n(S)} \\\\=\dfrac{79 }{80}\\\\=0.9875

4 0
2 years ago
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