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Kipish [7]
2 years ago
6

An account is opened with $7,595.96 with a rate of increase of 2% per year. After 1 year, the bank account contains $7,746.90. A

ssuming no deposits or withdrawals are made, which equation can be used to find y, the amount of money in the account after x years
Mathematics
1 answer:
wolverine [178]2 years ago
3 0

Answer:

y = 7,595.96(1.02)x

Step-by-step explanation:

y = 7,595.96(1.02)x

a normal linear equation is basically y = x + b

in this case,

  • y = amount of money in the bank account
  • x = number of years
  • b is not required because there are no new deposits or withdrawals from the account

Since the accounts earns 2% interest per year, then you need to multiply the original amount by 1.02. As more years pass, the account will increase by 1.02, so the account increases by 1.02x. Since the original amount is $7,595.96, that will be our starting point.

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Suppose you are considering making an investment. You have a chance to make an investment that will return either $50,000 or $10
timurjin [86]

Answer:

D because its rounded to the nearest ten

8 0
2 years ago
Read 2 more answers
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
Pablo wishes to grow his baseball card collection to at least 3000 cards. He currently has 1200 cards and knows that his favorit
tatuchka [14]

Answer:

B & E

Step-by-step explanation: They did it the right way it is supposed to be!

Brainiest would help! :D

5 0
2 years ago
3.12 Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in Ca
marusya05 [52]

Answer:

a) 93.943% = 93.9%

b) 93.528% = 93.5%

c) Speed of the fastest 5% ≥ 80.5 miles/hour

d) 29.46% = 29.5%

Step-by-step explanation:

Mean, xbar = 72.6 miles/hour.

standard deviation, σ = 4.78 miles/hour

For each of the questions, we'll need to normalize the speeds.

a) The standardized score for 80 miles/hour is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (80 - 72.6)/4.78 = 1.55

To determine the probability of a car having speed less than 80 miles/hour, P(x < 80) = P(z < 1.55)

We'll use data from the normal probability table for these probabilities

P(x < 80) = P(z < 1.55) = 1 - P(z ≥ 1.55) = 1 - P(z ≤ -1.55) = 1 - 0.06057 = 0.93943

b) percent of passenger vehicles travel between 60 and 80 miles/hour.

60 miles/hour standardized = (60 - 72.6)/4.78 = -2.64

We'll use data from the normal probability table for these probabilities

P(60 < x < 80) = P(-2.64 < z < 1.55) = P(z ≤ 1.55) - P(z ≤ -2.64) = 0.93943 - 0.00415 = 0.93528

c) How fast to do the fastest 5% of passenger vehicles travel?

We'll use data from the normal probability table for these probabilities

Top 5% corresponds to a z-score of 1.65. P(z ≥ 1.65) = 0.95053

1.65 = (x - 72.6)/4.78

x = 80.487 miles/hour = 80.5 miles/hour.

d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

70 miles/hour, standardized = (70 - 72.6)/4.78 = 0.54

P(x > 70) = P(z > 0.54) = 1 - P(z ≤ 0.54) = 1 - 0.7054 = 0.2946.

Hope this helps!!!!

6 0
2 years ago
Suppose a worker needs to process 100 items. The time to process each item is exponentially distributed with a mean of 1 minute,
Hunter-Best [27]

Answer: 0.0368

Step-by-step explanation:

The cumulative distribution function for exponential distribution is :-

P(x)=1-e^{-\lambda x}

, where \lambda is the mean of the distribution.

Given : The time to process each item is exponentially distributed with a mean of 1 minute .

In hour, the mean time to process each item = \lambda=\dfrac{1}{60} hour

Then , the probability that the worker finishes in less than 2.25 hours :-

P(x

Hence, the probability that the worker finishes in less than 2.25 hours = 0.0368

3 0
2 years ago
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