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inysia [295]
3 years ago
6

Mr. McKay spent $2.60 for a box of crackers and then divided them evenly between the s students in his classroom, Write an expre

ssion for the cost of crackers for each students. Find the cost for s = 20 students.
Mathematics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

You would go $2.60/s

Step-by-step explanation:

then when you have your total number of students you take the total cost and divide that by the number of students that you have to figure out the cost of crackers for each student

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if you place these marbles in a bag, close your eyes, and choose a marble, what is the probability that it will be
oksian1 [2.3K]

Answer:

P(blue) = 3/7

So '7' belongs in the box

Step-by-step explanation:

There are 14 total marbles, 6 of which are blue.  The probability of pulling a blue marble out of the bag is

P(blue) = 6/14, which reduces to 3/7

8 0
2 years ago
A study of the pricing accuracy of checkout scanners at a store found that one of every 20 items is priced incorrectly. On any g
GalinKa [24]

Answer:

a) There are going to be at least 15 items both priced correctly, and incorrectly, which means that we can use the normal approximation to the binomial to solve the exercise in this supposed problem.

b) 20.36% probability that exactly one of the five items is priced incorrectly by the scanner

c) 22.62% probability that at least one of the five items is priced incorrectly by the scanner

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is priced correctly, or it is not. The probability of an item being priced incorrectly is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Five items selected.

This means that n = 5

One of 20 is priced incorrectly.

This means that p = \frac{1}{20} = 0.05

a) Assuming that this store sells thousands of items every day, which probability distribution would you use and why?

There are going to be at least 15 items both priced correctly, and incorrectly, which means that we can use the normal approximation to the binomial to solve the exercise in this supposed problem.

b) What is the probability that exactly one of the five items is priced incorrectly by the scanner?

This is P(X = 1).

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

P(X = 1) = C_{5,1}.(0.05)^{1}.(0.95)^{4} = 0.2036

20.36% probability that exactly one of the five items is priced incorrectly by the scanner

c) What is the probability that at least one of the five items is priced incorrectly by the scanner?

Either no items are priced incorrectly, or at least one is. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 0

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

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

P(X = 0) = C_{5,0}.(0.05)^{0}.(0.95)^{5} = 0.7738

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7738 = 0.2262

22.62% probability that at least one of the five items is priced incorrectly by the scanner

5 0
3 years ago
If your really good with math pls help​
Amanda [17]

Answer:

GIVEN :

Which division expression is equivalent to 4 and one-third divided by Negative Start Fraction 5 over 6 End Fraction?

A) Start Fraction 13 over 3 End Fraction divided by (Negative Start Fraction 5 over 6 End Fraction)

B)  Negative Start Fraction 5 over 6 End Fraction divided by Start Fraction 13 over 3 End Fraction

C) Start Fraction 13 over 3 End Fraction divided by Start Fraction 5 over 6 End Fraction  

D) Negative Start Fraction 13 over 3 End Fraction divided by (Negative Start Fraction 5 over 6 End Fraction)

TO FIND :

The division expression is equivalent to 4 and one-third divided by Negative Start Fraction 5 over 6 End Fraction

SOLUTION :

Given expression is 4 and one-third divided by Negative Start Fraction 5 over 6 End Fraction.

It can be written as

Now solving the above expression as below :

By converting the mixed fraction into proper fraction,

By using the quotient rule of exponents :

∴ the given expression  is equivalent to

⇒ option A) Start Fraction 13 over 3 End Fraction divided by (Negative Start Fraction 5 over 6 End Fraction) is correct.

Step-by-step explanation:

7 0
3 years ago
Ten times the sum of half a number 6 is 8
Viefleur [7K]
=> 8 = 8 Maybe? 
Sorry if this isnt right!

7 0
3 years ago
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ELEN [110]

Answer:

3rd one have a great day

5 0
3 years ago
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