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Sav [38]
3 years ago
6

Find I. Choose the correct answer. Amount Financed (m) = $1,400 Number of Payments per year (y) = 12 Number of Payments (n) = 24

Total Interest (c) = $200.50
I =
Mathematics
2 answers:
kirill115 [55]3 years ago
8 0
Hi there!

The answer to your problem is I = 13.7%

Your friend, ASIAX
Ugo [173]3 years ago
3 0
If m is the amount financed and P is the periodic payment, m = P\cdot\dfrac{1 - v^{n}}{i}. This is the standard formula for periodic payment over time for level interest. You are given m = 1400 You are given n = 24 You are ALMOST given P. You can figure it out. P = (1400 + 200.50)/24 = 1600.50/24 = 66.69 You're almost done. There are two things yet to do.
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The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit
Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

b) We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean \mu, standard deviation \sigma

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

b. If the length of pregnancy is in the lowest a​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

8 0
3 years ago
Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex
Nataly [62]

Answer:

(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942  

(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between 0.0942 \leq p_A -p_B \leq 0.09978

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_A represent the real population proportion female for Biology

\hat p_A =\frac{84199}{144796}=0.582 represent the estimated proportion female for biology

n_A=144796 is the sample size for A

p_B represent the real population proportion female for calculus AB

\hat p_B =\frac{102598}{211693}=0.485 represent the estimated proportion female for Calculus AB

n_B=211693 is the sample size required for B

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942  

(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between 0.0942 \leq p_A -p_B \leq 0.09978

5 0
3 years ago
Lynne used 3/8 cup of flour and 1/3 cup of sugar in a recipe.what number is a common denominator for3/8 and 1/3
Liula [17]

Ahiffggggggggggggt was the morning I was going over the

7 0
3 years ago
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Which of the following is the value of pq ? 2x+41
loris [4]

m pq = 2 Have a good day! sorrey if i'm wrong

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Surface area<br> Pleases help ASAP!!!
PilotLPTM [1.2K]
2(20 x 10) + 2(10 x 5) + 2(20 x 5)
2(200) + 2(50) + 2(100)
400 + 100 + 200
400 + 300
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