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puteri [66]
3 years ago
10

Brandon has a box 9.5 inches tall, 3.8 inches wide, and 4.2 inches long. How many square inches of wrapping paper does he need t

o wrap the box?
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

183.92 inch²

Step-by-step explanation:

To find the total amount of wrapping paper needed, you must find the Surface Area of the box.

Surface area of a rectangular prism is: A = 2(<em>wl</em> + <em>hl</em> + <em>hw</em>)

w = width = 3.8

h = height = 9.5

l = length = 4.2

Plug in the corresponding numbers to the corresponding variables.

A = 2(<em>wl</em> + <em>hl </em>+ <em>hw</em>)

A = 2((3.8 * 4.2) + (9.5 * 4.2) + (9.5 * 3.8))

Simplify.

A = 2((15.96) + (39.9) + (36.1))

A = 2(15.96 + 39.9 + 36.1)

A = 2(91.96)

Multiply.

A = 183.92

183.92 inch² is your answer.

~

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The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.
Mama L [17]

Answer:

a) P(Y > 76) = 0.0122

b) i) P(both of them will be more than 76 inches tall) = 0.00015

   ii) P(Y > 76) = 0.0007

Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

                    the probability that he will be more than 76 inches tall.

              (b) If two men are chosen at random from the population, find

                    the probability that

                    (i) both of them will be more than 76 inches tall;

                    (ii) their mean height will be more than 76 inches.

Proof -

a)

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{S.D}) > \frac{( 76- mean)}{S.D})

                 = P(Z >  \frac{( 76- mean)}{S.D})

                 = P(Z > \frac{76 - 69.7}{2.8})

                 = P(Z > 2.25)

                 = 1 - P(Z  ≤ 2.25)

                 = 0.0122

⇒P(Y > 76) = 0.0122

b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

\frac{S.D}{\sqrt{N} } = 1.979899,

Now,

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{\frac{S.D}{\sqrt{N} } })) > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- 69.7)}{1.979899 }))

                 = P(Z > 3.182)

                 = 1 - P(Z ≤ 3.182)

                 = 0.0007

⇒P(Y > 76) = 0.0007

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