All categories

3 years ago

if serena invested the 2500 in the cd that she yields at 4% interest, what will the cdbe worthafte 2 years?

Mathematics

2 answers:

Svetlanka [38]3 years ago

8 0

Your answer would be 2500

Naya [18.7K]3 years ago

6 0

Your answer would be <u>2700</u>

Using I=P*r*t

P= 2500

R= 4% (substitue to 0.04)

T= 2 Years

I= 2500* 0.04 * 2

I= 100* 2

I= 200

2500+200=

2700

You might be interested in

Area and percent .

siniylev [52]

Answer:

a. 61.92 in²

b. 21.396 ≈ 21.4%

c. $4.71

Step-by-step explanation:

a. Amount of waste = area of rectangular piece of stock - area of two identical circles cut out

Area of rectangular piece of stock = 24 in × 12 in = 288 in²

Area of the two circles = 2(πr²)

Use 3.14 as π

radius = ½*12 = 6

Area of two circles = 2(3.14*6²) = 226.08 in²

Amount of waste = 288 - 226.08 = 61.92 in²

b. % of the original stock wasted = amount of waste ÷ original stock × 100

= 61.92/288 × 100 = 6,162/288 = 21.396 ≈ 21.4%

c. 288 in² of the piece of stock costs $12.00,

Each cut-out circle of 113.04 in² (226.08/2) will cost = (12*113.04)/288

= 1,356.48/288 = $4.71.

3 0

3 years ago

If numerator is 2 less than the denominator of a rational number and when 1 is subtracted from numerator and denominator both, t

LUCKY_DIMON [66]

Answer:

3/5

Step-by-step explanation:

1/2=2/4

2+1=3

1+4=5

3/5

6 0

2 years ago

How I can answer this question, NO LINKS, if you answer correctly I will give u brainliest!

attashe74 [19]

Answer

Reson

Total = 21

21/7 = 3

2 x 3 = 6

Only punkin plants so

it's B.

6 0

2 years ago

Read 2 more answers

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

Given the function f(x)=3x-4/5 which of the below expressions is correct?

natta225 [31]

Is it an inverse function

4 0

3 years ago