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Dvinal [7]
3 years ago
8

Mary has 50 earring boxes. The boxes measure 1 inch on each side. How many rectangular prisms, each with a different – sized bas

e, could Mary make by stacking all of the earring boxes?
A) 10

B)15 PLZZZ HELP!!!

C)25

D)50
Mathematics
2 answers:
Westkost [7]3 years ago
8 0
Answer:D)50

WORK: 

Well box is 1 inch by 1 inch so thats 1x50=50 

So answer is 50 :33

**Sorry if wrong tried my best :33*8


kvasek [131]3 years ago
3 0

Answer with explanation:

Number of earring boxes that Mary has =50

Dimension of each box = 1 inch× 1 inch × 1 inch=1 cubic inches

⇒Number of rectangular prisms, each with a different – sized base, that Mary can make by stacking all of the earring boxes

                     =Then, the Dimension of Box will be as follows

         = [ Length=1 inch , 1 inch≤Base≤50 inch, Height =1 inch]

=[(1,1,1), (1,2,1),(1,3,1),(1,4,1),(1,5,1),(1,6,1),(1,7,1),(1,8,1),(1,9,1),(1,10,1),(1,11,1),(1,12,1),.....,(1,50,1)]

=So, there will be , 50 different Rectangular prism ,earrings.

⇒⇒There is another way of looking at the Question.If you have to ,use each earring once then,Number of rectangular prism each with different size

      =[(1,1,1), (1,2,1),(1,3,1),(1,4,1),(1,5,1),(1,6,1),(1,7,1),(1,8,1),(1,9,1)], 5 earrings will be left,when stacked together will form the combination in the set Provided.

→So, total number of Distinct earring = 9 Rectangular prisms

Otherwise,→ number of Distinct Earring =50

Option D: 50

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Step-by-step explanation:

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Answer:

Probability that the student scored between 455 and 573 on the exam is 0.38292.

Step-by-step explanation:

We are given that Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118.

<u><em>Let X = Math scores on the SAT exam</em></u>

So, X ~ Normal(\mu=514,\sigma^{2} =118^{2})

The z score probability distribution for normal distribution is given by;

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Now, the probability that the student scored between 455 and 573 on the exam is given by = P(455 < X < 573)

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       P(X < 573) = P( \frac{X-\mu}{\sigma} < \frac{573-514}{118} ) = P(Z < 0.50) = 0.69146

       P(X \leq 2.9) = P( \frac{X-\mu}{\sigma} \leq \frac{455-514}{118} ) = P(Z \leq -0.50) = 1 - P(Z < 0.50)

                                                         = 1 - 0.69146 = 0.30854

<em>The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.69146.</em>

Therefore, P(455 < X < 573) = 0.69146 - 0.30854 = <u>0.38292</u>

Hence, probability that the student scored between 455 and 573 on the exam is 0.38292.

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