Answer:
m(x) = 5 + 100x
Step-by-step explanation:
Let m(x) be the balance of the account after x months.
m(x) = 5 + 100x
The expression is equivalent to 1.5 raised to the fifth power divided by 0.7 raised to the fourth power, all raised to the third power is; Option A; 1.5 raised to the fifteenth power divided by 0.7 raised to the twelfth power
<h3>How to solve exponents?</h3>
We want to find the expression that is equivalent to 1.5 raised to the fifth power divided by 0.7 raised to the fourth power, all raised to the third power.
The sentence above can be expressed as;
((1.5⁵)/(0.7⁴))³
This can be broken down into;
[1.5^(5 * 3)]/(0.7^(4 * 3))
1.5¹⁵/0.7¹²
Looking at the options, the only correct one is;
Option A; 1.5 raised to the fifteenth power divided by 0.7 raised to the twelfth power
Read more about Exponents at; brainly.com/question/11975096
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Answer:
solution
Step-by-step explanation:
solution is the answer
The Decimal number system has following rules.
Suppose you have to write a number 4,567,892 .
2 -ones - (first period)
9-tens - (first period)
8-hundreds - (first period)
7-thousands - (second period)
6-ten thousand- (second period)
5-hundred thousand- (second period)
4-million- (third period)
then comes ten million,hundred million and then Billion.
If a number is given you have to apply commas from left to right after every three places.
So, your number is 9,418.
8-first period
1--first period
4--first period
9- second period
Answer:
a) the probability that the minimum of the three is between 75 and 90 is 0.00072
b) the probability that the second smallest of the three is between 75 and 90 is 0.396
Step-by-step explanation:
Given that;
fx(x) = { 1/5 ; 50 < x < 100
0, otherwise}
Fx(x) = { x-50 / 50 ; 50 < x < 100
1 ; x > 100
a)
n = 3
F(1) (x) = nf(x) ( 1-F(x)^n-1
= 3 × 1/50 ( 1 - ((x-50)/50)²
= 3/50 (( 100 - x)/50)²
=3/50³ ( 100 - x)²
Therefore P ( 75 < (x) < 90) = ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx
= 3/50³ [ -2 (100 - x ]₇₅⁹⁰
= (3 ( -20 + 50)) / 50₃
= 9 / 12500 = 0.00072
b)
f(k) (x) = nf(x) ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k
Now for n = 3, k = 2
f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))
= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)
= 6/50³ ( 150x - x² - 5000 )
therefore
P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx
= 99 / 250 = 0.396
