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White raven [17]

3 years ago

13

Ieda bakes 73 cupcakes and accidentally drops 5 of them. She divides them equally into 13 containers for a bake sale for the sch

ool band. How many cupcakes does Ieda have left?
2 cupcakes
3 cupcakes
4 cupcakes
5 cupcakes

2 answers:

Norma-Jean [14]3 years ago

7 0

3 cupcakes:
When 5 drops, the remaining are;
73-5=68 cupcakes
13 containers will hold 65 cupcakes

Scrat [10]3 years ago

4 0

If Leda baked 73 cupcakes and accidentally drops 5. She has 68 left.

73 - 5 = 68

If she divides them in equal container she would have 3 left.

13 x 5 = 65
Therefore: 68/15= 5 r 3

Therefore Leda had 5 cupcakes in each container and 3 left.

Answer: 3 cupcakes

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3 years ago

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Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

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$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

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b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

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We have to find the value of x such that the probability is 0.85.

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$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

Calculation the value from standard normal z table, we have,

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

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