All categories

1 year ago

Why are 1995 pennies worth less than 2018 pennies?

Mathematics

1 answer:

Gekata [30.6K]1 year ago

7 0

Answer:

Most 1995 pennies in circulated condition are only worth their face value of $0.01. These coins can only sell for a premium in uncirculated condition. The 1995 penny with no mint mark and the 1995 D penny are each worth around $0.30 in uncirculated condition with an MS 65 grade.

Step-by-step explanation:

hope this helps.

You might be interested in

Please help me complete this !! 12points

Svet_ta [14]

Answer:

x1=1

x2=3

y1=5

y2=1

midpoint= (x1+x2/2, y1+y2/2)=(1+3/2,5+1/2)=(2,3)

x1=2

x2=4

y1=4

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(2+4/2,4+6/2)=(3,5)

x1= -2

x2=4

y1=4

y2= -6

midpoint= (x1+x2/2, y1+y2/2)=(-2+4/2,4-6/2)=(1,-1)

x1=0

x2=0

y1=3

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(0/2,3+5/2)=(0,4)

x1= -2

x2=2

y1= -6

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(-2+2/2,-6+6/2)=(0,0)

x1=1

x2=4

y1=4

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(1+4/2,4+5/2)=(5/2,9/2)

x1=-3

x2=-4

y1=5

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(-3-4/2,5+5/2)=(-7/2,5)

x1=5

x2= -4

y1=2

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(5-4/2,2+6/2)=(1/2,4)

3 0

2 years ago

What number can you use to simplify 5 and 15?

pshichka [43]

Answer:

You can use 5 to simplify 5 and 15, giving you 1 and 3

If this answer was helpful please consider giving brainliest!

7 0

2 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c: