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2 years ago

13

You saved $35 to buy two books. The first book cost an unknown amount and the second book cost $24. If you don't have any money

left over, how much did the first book cost?

1 answer:

vovangra [49]2 years ago

5 0

Answer:

$11 Dollar

Step-by-step explanation:

35 - 24 = 11

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Carol makes 9 1/3 cups of snack mix.She puts 2/3 all the snack mix into plastic bags. She puts of a cup of the snack mix in each

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Carol has 6 2/9 cups of snack mix to put into plastic bags, so she would need 6 plastic bags and she would have 2/9 of snack mix left over.

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2 years ago

Describe how you figure out the square footage of the walls in a 10 by 14 room with an 8 foot ceilings.

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2 years ago

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.

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

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3 years ago

How do you solve this

inessss [21]

Answer:

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Step-by-step explanation:

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2 years ago

Assume every 10-digit natural number is a possible telephone number except those that begin with 0, 1 or 2. What fraction of tel

lana [24]

Answer: 1/70

Step-by-step explanation:

This is a question that can also be interpreted as what is the probability of having the first number of a phone number to be 8 and the last number of the phone number to also be 8. This answer gives the fraction of the phone numbers that starts with 8 and end with 8.

Since three numbers (0,1,2) cannot start a phone number and we are left to pick from 7 numbers,

then the probability of figure "8" starting phone number = 1/7

Since all 10 numbers can possibly end a phone number,

then the probability of having figure "8" as the last digit of a phone number = 1/10

Hence probability of having "8" as the first and last digit of a phone number = fraction of total telephone numbers that begin with digit 8 and end with digit 8 = 1/7 × 1/10 = 1/70.

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2 years ago