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Vsevolod [243]
3 years ago
8

3200 dollars is placed in an account with an annual interest rate of 6.5%. To the nearest year, how long will it take for the ac

count value to reach 15300 dollars?
Mathematics
1 answer:
madam [21]3 years ago
5 0

4 yrs

Step-by-step explanation:

Formulae for simple interest is;

A= P (1 + rt)                Whereby;

   A = Total Accrued Amount (principal + interest)

   P = Principal Amount

   I = Interest Amount

   r = Rate of Interest per year in decimal; r = R/100

   R = Rate of Interest per year as a percent; R = r * 100

   t = Time Period involved in months or years

15300 = 3200 (1 + 106.5/100 *t)

15300/3200 = 1 + 1.065t

4.78125 – 1 = 1.065t

3.78125 = 1.065t

3.78125/1.065 = t

3.55 = t

Rounded off to the nearest whole number;

= 4

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mr Goodwill [35]

The sum of angles of a triangle is 180°, so m∠K = 180° -45° -30° = 105°.

The Law of Sines tells you

... FL/sin(∠K) = FK/sin(∠L)

Solving for FL, we get

... FL = FK·sin(∠K)/sin(∠L)

... FL = a·sin(105°)/sin(30°) = a·sin(105°)/(1/2)

... FL = 2a·sin(105°) ≈ 1.93185a

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3 years ago
On your wedding day you leave for the church 34 minutes before the ceremony is to begin, which should be plenty of time since th
Schach [20]

Answer:

Average speed   is 37.35 mi/h

Step-by-step explanation:

given data

leave =  34 minutes before

church distance =  12.0 miles

average speed first 17 minutes =  5.0 mi/h

solution

so we find Total distance travel in first 17 minutes = speed × time

Total distance travel in first 17 minutes = 5 × \frac{17}{60}

Total distance travel in first 17 minutes = 1.416 mi

and

Distance Remaining = 12 - 1.416 = 10.584 mi

Time Remaining = 34 - 17 min = 17 min

so

remaining distance Average speed  = \frac{distance}{time}

Average speed  = \frac{10.584}{\frac{17}{60}}

Average speed   is 37.35 mi/h

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A television that listed for $550 was on sale for 35% discount. What is the sale price?
katrin [286]
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3 0
3 years ago
Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance o
AysviL [449]

Answer:

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

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.

Mean 490 mg and variance of 400.

This means that \mu = 490, \sigma = \sqrt{400} = 20

What is the probability that a randomly selected pill contains at least 500 mg of minerals?

This is 1 subtracted by the p-value of Z when X = 500. So

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

Z = \frac{500 - 490}{20}

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

4 0
3 years ago
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