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Ludmilka [50]
3 years ago
9

Which on is the answer? Pls help

Mathematics
2 answers:
zheka24 [161]3 years ago
4 0
Both is the answer to the problem
blagie [28]3 years ago
3 0

If I'm not mistaking I think the answer is both

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I NEED HELP PLEASE, THANKS! :) Heather invests $4,900 in an account with a 3.5% interest rate, making no other deposits or withd
Alex

Answer:    $5,828.28

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

Use the Compound Interest formula:     A=P(1+\frac{r}{n})^{nt}       where

  • A is the accrued amount (balance)
  • P is the principal (initial amount invested)
  • r is the interest rate (in decimal form)
  • n is the number of times compounded each year
  • t is the time of the investment (in years)

Given: P = 4,900

           r = 3.5% (0.035)

           n = 2

           t = 5

A=4,900\bigg(1+\dfrac{0.035}{2}\bigg)^{2(5)}\\\\\\.\qquad = 4900(1.0175)^{10}\\\\\\.\qquad = \large\boxed{5,828.28}\\

5 0
3 years ago
There are 47 numbers Find the probability that i will pick the 6 winning numbers in a draw​
Agata [3.3K]

Answer: You have six numbers on your ticket. In drawing the first number from the collection of 45, one of those six numbers must be drawn in order for you to still have a chance of winning the big prize. So following the draw, the probability is only 6/45 (or 1/9) that you’re still in the game.

You can now see that on each successive draw the probability steadily drops- to 5/44, 4/43, 3/42, 2/41, and for the last number 1/40. As other answers have stated, there are 8,145,060 possible draws of six numbers and only one makes you a winner.

If you bought ten tickets for every drawing, the odds reach a 50% chance of having won at about 520,550 drawings. Assuming two drawings per week, the chance of winning reaches 50% in about 5,200 years.

Of course, SOMEONE will win and if you don’t play, your chance of winning is zero. It’s kind of fun to occasionally buy lottery tickets just for the dream of winning but don’t spend a lot of money on them.

5.9K viewsView upvotes · Answer requested by Arthur C. Thorpe

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