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4 years ago

8

In june 2005 peter mailed a package from his local post office in fayetteville north carolina to a friend in radford virginia fo

r 2.07 the first class rate at the time was 0.23 per once

1 answer:

erik [133]4 years ago

7 0

0.23w = 2.07

w = 2.07 ÷ 0.23

w = 9 ounces

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abruzzese [7]

Answer:

its 1/4 :)

Step-by-step explanation:

8 0

3 years ago

Find the balance on a deposit of $1275 that earns 9% simple interest for 3 years.

never [62]

Answer:

<em>B) $1619.25</em>

Step-by-step explanation:

<em>Equation: </em>

<em>A = P(1 + rt) </em>

<em> </em>

<em>Calculation: </em>

<em>First, converting R percent to r a decimal </em>

<em>r = R/100 = 9%/100 = 0.09 per year. </em>

<em> </em>

<em>Solving our equation: </em>

<em>A = 1275(1 + (0.09 × 3)) = 1619.25 </em>

<em>A = $1,619.25 </em>

<em> </em>

<em>The total amount accrued, principal plus interest, from simple interest on a principal of $1,275.00 at a rate of 9% per year for 3 years is $1,619.25.</em>

Hope this helps, have a good day. c;

6 0

3 years ago

Read 2 more answers

The store bought a table for $150, and sold it for $240. What percentage was the markup?

Akimi4 [234]

Answer:50%

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

12-(4x2)=(12-4)x(12-2)=

faust18 [17]

Answer:

The awnser is Incorrect

Step-by-step explanation:

So do the Parentheses first (4x2)=8

12-4=8

12-2=10

Now do your multiplcation 8x10=80

Subtract 12-8=4

So the awnser is incorrect

6 0

3 years ago

Read 2 more answers

An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

5 0

3 years ago