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dexar [7]
2 years ago
15

LeAnn is purchasing giftwrap for a box that measures 8 inches long, 6 inches wide, and 6 inches tall. Calculate the total area o

f the box that will be covered in gift wrap. Show all of your work for full credit.
Mathematics
2 answers:
Pavel [41]2 years ago
8 0

Let

L---------> the length side of the box

W--------> the width side of the box

H-------> the height of the box

we know that

L=8\ in\\W=6\ in\\H=6\ in

the surface area of the box is equal to

S=2*area\ of\ the\ base+perimeter\ of\ the\ base*H

<u>Find the area of the base</u>

A=L*W=8*6=48\ in^{2}

<u>Find the perimeter of the base</u>

P=2*L+2*W=2*8+2*6=28\ in

<u>Find the surface area</u>

S=2*48+28*6

S=264\ in^{2}

therefore

<u>the answer is</u>

the total area of the box that will be covered in gift wrap is 264\ in^{2}

sertanlavr [38]2 years ago
5 0
A box shape in three dimensional space. Formally, a polyhedron for which all faces are rectangles. 
Rectangular Parallelepiped
Volume = lwh 8*6*6=288Lateral Surface Area = 2lh + 2wh=94Surface Area = 2lw + 2lh + 2wh=288
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N = original number = 6
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Three students A, B, and C are enrolled in the same class. Suppose that A attends class 30 percent of the time, B attends class
Dvinal [7]

Answer:

(a) 0.93

(b) 0.38

Step-by-step explanation:

A attends class 30% of the time, so DO NOT attend 70%

B attends class 50% of the time, so DO NOT attend 50%

C attends class 80% of the time, so DO NOT attend 20%

Writing as probabilities:

P(A) = 0.30 and P(A') = 0.70

P(B) = 0.50 and P(B') = 0.50

P(C) = 0.80 and P(C') = 0.20

(a) the probability that at least one of them will be in class on a particular day

Let's call Q the event of none of them be in class on a particular day

Probability of at least one be in class is the complement of none of them be there, so: 1 - P(Q)

P(Q) = 0.7*0.5*0.2 = 0.07

1 - P(Q) = 1 - 0.07 = 0.93

(b) the probability that exactly one of them will be in class on a particular day?

One of them exactly be in class is

A is B not C not or A not B is C not or A not B not and C is, so

P(A).P(B').P(C') + P(A').P(B).P(C') + P(A').P(B').P(C)

0.3*0.5*0.2 + 0.7*0.5*0.2 + 0.7*0.5*0.8 =

0.03 + 0.07 + 0.28 = 0.38

4 0
2 years ago
Find the area for a trapezoid 9.85 m 15 m 11 m 9 m
Otrada [13]

Answer:

total area = 99m² + 39.4m²

= 138.4m²

7 0
2 years ago
Sue either travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4 . The
Brilliant_brown [7]

Answer: P(A∪B)=0.72. The answer is provided below.

Step-by-step explanation:

Sue travels by bus or walks whenever she goes to the shops.

Probability of (catching the bus to shop ), P(A)= 0.4

Probability of (catching the bus from the shop ), P(B) = 0.7

Both of the events A and B are independent events.

Therefore, P(A∩B) = 0.4 × 0.7

 = 0.28

Probability that Sue walks one way will be = 1 - P(A∩B)

= 1 - 0.28

= 0.72

Hence, the probability that Sue will walk at one way is 0.72

7 0
2 years ago
Given that lim x→a f(x) = 0 lim x→a g(x) = 0 lim x→a h(x) = 1 lim x→a p(x) = ∞ lim x→a q(x) = ∞, evaluate the limits below where
love history [14]
Lim x→a f(x) = 0
lim x→a g(x) = 0
lim x→a h(x) = 1
lim x→a p(x) = ∞
lim x→a q(x) = ∞

(a) lim x→a [f(x) − p(x)]
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0 - ∞
-∞

(b) lim x→a [p(x) − q(x)]
lim x→a p(x) - lim x→a q(x)
∞ - ∞
0

(c) lim x→a [p(x) + q(x)]
lim x→a p(x) + lim x→a q(x)
∞ + ∞
∞

3 0
2 years ago
Read 2 more answers
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