To solve this, we are going to use the compound interest formula:

where

is the final amount after

years

is the initial investment

is the interest rate in decimal form

is the number of times the interest is compounded per year
For the first 4 years we know that:

,

,

, and since the problem is not specifying how often the interest is communed, we are going to assume it is compounded annually; therefore,

. Lest replace those values in our formula:




Now, for the next 6 years the intial investment will be the final amount from our previous step, so

. We also know that:

,

, and

. Lets replace those values in our formula one more time:




We can conclude that Collin will have <span>£3691.41 in his account after 10 years.</span>
3x23=69 is the equation that has the sum 69
Answer:
1) 120
2) E (Z) = 12 and Variance of Z = 90
a) 5 liters
Step-by-step explanation:
1. We can find this by suing combinations.
Here n= 10 and r= 3 so n C r
= 10 C 3= 120
2. E(X) = 8 and E(Y) = 3
Z = 2X - 3Y +5
E(Z ) = 2 E (X) - 3(E)(Y) +5 ( applying property for mean)
= 2(8) - 3(3)+ 5 = 16+5-9= 21-9= 12
V(X) = 9 and V(Y) = 6.
V(Z) = E(Z )²- V(X) *V(Y) (applying property for Variance for two variables )
= 144- 54= 90
3. 55 liters contain adulterated milk in 7: 4.
So it contains 4/ 11*55= 20 liters of water
But we want to make it a ratio of 7:6
the water will be 6/13 *55= 25.38 when rounded gives 25 liters of water
So 25- 20 = 5 liters must be added to make it a ratio of 7:6
Answer:
10 cm
Step-by-step explanation:
Given:
No. of small spherical bulb = 1,000
radius (r) of smaller bulbs = 1 cm
Required:
radius of the bigger bulb
SOLUTION:
The following equation represents the relationship of the volume of the smaller and bigger bulb,

Where,
= volume of bigger bulb
= volume of smaller bulb
1,000 is the number of smaller bulbs melted to form the bigger bulb.
Volume of a sphere is given as, ⁴/3πr³
Therefore:
= ⁴/3*π*r³ = 4πr³/3
= ⁴/3*πr³ = ⁴/3*π*(1)³ = ⁴/3π*1 = 4π/3
Plug the above values into the equation below:





(12pie cancels 12 pie)
(taking the cube root of each side)
Radius of the bigger bulb = 10 cm