In three more years, Angelo’s grandfather will be six times as old as Angelo was last year. When

A bank account yields 7 percent interest, compounded annually. If you deposit $1,000 in the account, what will the account balan

lianna [129]

Hi there
The formula is
A=p (1+r)^t
A future value?
P present value 1000
R interest rate 0.07
T time 5 years
So
A=1,000×(1+0.07)^(5)
A=1,402.55

It's a

Hope it helps

7 0

3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

Item 12 You win an online auction for a toy. Your winning bid of $52 is 65% of your maximum bid. How much more were you willing

nataly862011 [7]

Answer:

$28

Step-by-step explanation:

Given that:

Value of Winning bid = $52

Winning bid 65% of the maximum bid.

To find:

How much more is the maximum bid from the winning bid ?

Solution:

We are given that the winning bid is 65% of the maximum bid.

Using this percentage value, we need to first find the value of maximum bid and then we need to subtract the value of winning bid from the maximum bid to find the answer.

Let the value of maximum bid = $ $x$