If I'm putting £100 a month into a fund that has 13.03% annual interest annually for 48 years, what's the the total?

Determine whether the statement is true or false. If it is true, explain why.

Over [174]

True.

Since $y^4 \geq 0$ , $y' = -1 - y^4 \ \textless \ 0$ and the solutions are decreasing functions.

8 0

3 years ago

In order to get an A in science, Jack must get at least a 92% on his next test. If test has 30 questions,what is the minimum num

Neko [114]

Jack must score 28 problems correct. To get a 92% he must answer 92% of the 30 questions right. which means 0.92*30=27.6, but it is not possible to score 0.6 of a problem right so you would round to 28, which guarantees Jack scores a 92 because if we round down, Jack would score less than a 92

7 0

3 years ago

Read 2 more answers

Todd is 81 years younger than Rosa. 3 years ago, Rosa’s age was 4 times Todd’s age. How old is Todd now?

Delvig [45]

Let t represent Todd's age now.
.. 4(t -3) -(t -3) = 81 . . . . . . 3 years ago, their differnce in ages was 81.
.. 3t -9 = 81
.. t = (81 +9)/3 = 30

Todd is 30 now.

_____
You can also work this by considering "ratio units." 3 years ago, the ratio of their ages was 4:1, a difference of 3. That difference corresponds to 81 years, so each "ratio unit" represents 81/3 = 27 years. Todd's age then was 1 ratio unit, 27 years. Now, Todd's age is 30.

8 0

3 years ago

PLZ Hurry!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!The sum of two numbers is 19 and their difference is 7, What are the two num

yuradex [85]

Answer:

13 and 6

Step-by-step explanation:

4 0

2 years ago

How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

ryzh [129]

<u>Answer</u>:

3003 number of 5-member chess teams can be chosen from 15 interested players.

<u>Step-by-step explanation:</u>

Given:

Number of the interested players = 15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula $nCr = \frac{n!}{r!(n - r)!}$

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now we have n = 15 and r = 5

Substituting the values,

$15C_5 = \frac{15!}{5!(15- 5)!}$

$15C_5 = \frac{15!}{5!(10)!}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11}{(5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{360360}{120}$

$15C_5 = 3003$

7 0

3 years ago