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Maru [420]
2 years ago
5

Nicolas has $650 to deposit into two different savings accounts. • Nicolas will deposit $400 into Account I, which earns 3.5% an

nual simple interest. • He will deposit $250 into Account II, which earns interest compounded annually. Nicolas will not make any additional deposits or withdrawals. Which amount is closest to the total balance of these two accounts at the end of 2 years? A $672.13 B $695.00 C $694.25 D $694.51 Mathematics
Mathematics
1 answer:
Nitella [24]2 years ago
6 0
The initial amount is $650. The $400 is deposited in Account 1 with 3.5% annual simple interest. The $250 is deposited in Account 2 with 3.5% interest compounded monthly.
n = 2 years
Account 1: Simple Interest

F = P (1+rn)
   = 400 (1 +(0.035*2))
   = $428

Account 2: Compound Interest

F = P (1+i)^n
   = 250 ( 1+0.035)^2
F =  267.81

Total money after 2 years = $428 + $267.81
 T = $695.81
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the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of
soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = \sigma = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

z=\frac{x-u}{\sigma}

Using the values, we get:

z=\frac{25-22}{4}=0.75

So,

P(X ≥ 25) is equivalent to P(z ≥ 0.75)

Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20

Thus, a test score of 20 represent the 31st percentile of the distribution.

3 0
2 years ago
A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th
GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.22, n = 425

So

\mu = E(X) = np = 425*0.22 = 93.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

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

Z = \frac{82 - 93.5}{8.54}

Z = -1.35

Z = -1.35 has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

8 0
2 years ago
Rob says that this graph shows a proportional<br> relationship. Is he correct? Explain.
zysi [14]

Answer:

No. Rob was correct to extend the lines to the origin, but the points should be on a single straight line. He made two straight lines.

Step-by-step explanation:

this is from the sample responds so you can see what they want

5 0
2 years ago
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Jude has 5 pairs of sunglasses that cost the same in his online shopping cart, then decides to get only 2. Each pair of sunglass
hoa [83]

Answer:

Step-by-step explanation:

5p is the original price

Then 2p is the new price

5p-2p=3p

5 0
3 years ago
A symmetrical binomial distribution with large sample size can be approximated by a __________
sweet [91]
Typical distribution?
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2 years ago
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