What is a number in which the digit 7 is worth 70

John deposited $2860 in a bank that pays 9% interest, compounded monthly. find the amount he will have at the end of 3 years ?

7nadin3 [17]

To find that you first need to find what 9% is. You can multiply the number, 2,860 by .09 to find out your answer. That should be equal to 257.40. Since you now have that number, you can multiply it by 36. 257.40 is your average per month and you want to know by 3 years. So then your next step is 257.4*36 (there is 36 months in 3 years) which should be equal to 9,266.40.

4 0

3 years ago

Read 2 more answers

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

natulia [17]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago

HELP ASAP Use the data below

beks73 [17]

Answer:

Part A:

-Minimum: 10

-Q1: 17.5

-Median: 30

-Q3: 42.5

-Maximum: 50

Step-by-step explanation:

Part B: IQR= 25

This shows that the data varies for 25 different numbers. That HALF of the data is between 25 numbers.

Part C: Using a box-and-whisker plot you can interpret the different values. Minimum is the first dot (10), connected to the first line (Q1 which is 17.5), connected by a box to the median (30), connected by a box to the third line (Q3 which is 42.5), connected to the last dot which is the maximum (50). And IQR is Q3-Q1, so 42.5-17.5 which is 25.

8 0

4 years ago

Find the sumofthe geometrical progression of five terms, of which the first term is 7 and the multiplier is 7.Verify that the su

777dan777 [17]

Answer:

The sum of first five term of GP is 19607.

Step-by-step explanation:

We are given the following in the question:

A geometric progression with 7 as the first term and 7 as the common ration.

$a, ar, ar^2,...\\a = 7\\r = 7$