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Tomtit [17]
1 year ago
6

Lindy is having a bake sale. She has 48 chocolate chip cookies to put in bags. How many bags can she fill if she puts the same n

umber in each bag and uses them all? Find all the possibilities. Explain your reasoning.
Mathematics
2 answers:
kogti [31]1 year ago
7 0
<span>1, 2, 3, 6, 8, 12, 16, 24, 48.
The number of bags she can make are shown in the table:<span><span>Number of CC cookies in each bag12346812162448</span><span>Number of bags48241612864321</span></span>Since the number of cookies that goes in a bag is a factor of the total number of cookies, all the factors of 48 are listed in the first row of the table. Each column corresponds to a situation where the 48 cookies are divided equally among some bags, and the product of the numbers in each column is 48. The number of bags is also a factor of the total number of cookies.</span>
spayn [35]1 year ago
4 0
Well how many bags? I would say find the factors of 48. You could do four bags with twelve cookies each. Or six bags with eight cookies each.
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For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
aleksley [76]

Answer:

a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

Option c) .253 is.correct

c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

= 0.344

Option c is.correct

d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

7 0
2 years ago
In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

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

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
1 year ago
Sonny deposited $8,500 in an account that earns 4% simple interest. If Sonny makes no more deposits or withdrawals, then how muc
SashulF [63]

Answer: $680.00

Step-by-step explanation:

Principal= 8500

Rate= 4%

Time= 2 years

Interest= (Principal×Rate×Time)/100

= (8500×4×2)/100

= 68000/100

= $680

The interest is $680

5 0
2 years ago
Read 2 more answers
Jenna dances for 3 hours on sunday, 2 hours on Monday and Tuesday, 1 hour on THursday,1.5 hours oon friday, and 2 hours on Satur
Delicious77 [7]

Answer: The average number of hours she danced per day is 1.9 hours (rounded to the nearest tenth)

Step-by-step explanation: We start by calculating how many hours she danced all together which can be derived as follows;

Summation = 3 +2 +2 + 1 + 1.5 + 2 = 11.5

The number of days she danced which is the observed data is 6 days (she did  not dance at all on Wednesday).

The average (or mean) hours she danced each day can be calculated as

Average = ∑x ÷ x

Where ∑x is the summation of all data and x is number of observed data

Average = (3+2+2+1+1.5+2) ÷ 6

Average = 11.5 ÷ 6

Average = 1.9166

Approximately, average hours danced is 1.9 hours (to the nearest tenth)

8 0
1 year ago
Sarah and thomas are going bowling. the probability that sarah scores more than 175 is 0.4, and the probability that thomas scor
pychu [463]

When the occurrence of one event say A does not affect the occurrence of another event say B, than the two events are said to be independent such that;

\\  &#10;P(A\cap B)=P(A)\times P(B)

where, P(A) = probability of occurrence of event A

and P(B) = probability of occurrence of event B

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Now, let event A = Sarah scores more than 175

and event B = Thomas scores more than 175

Thus, P(A)= Probability that Sarah scores more than 175 = 0.4

and P(B)= Probability that Thomas scores more than 175 = 0.2

Since, the scores are independent, thus the probability that both Sarah and Thomas scores more than 175 is,

\\  &#10;P(A\cap B)=P(A)\times P(B)\\  &#10;P(A\cap B)= 0.4\times 0.2= 0.08\\

Hence, the required probability is 0.08

(b).

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\\  &#10;P(A\cap B)=P(A)\times P(B\setminus A)\\

Now, let event A = Sarah scores more than 175

and event B = Thomas scores more than 175

Thus, P(A)= Probability that Sarah scores more than 175 = 0.4

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and P(B|A) = Sarah scores more than Thomas given that Thomas scores more than 175 = 0.3

Thus, the required probability is calculated as follows;

\\  &#10;P(A\cap B)=P(A)\times P(B\setminus A)\\  &#10;P(A\cap B)=0.2\times 0.3=0.06



             



             


7 0
1 year ago
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