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maria [59]
2 years ago
12

Find the area of the shaded region.

Mathematics
1 answer:
SCORPION-xisa [38]2 years ago
8 0

Answer:

C. 96 in²

Step-by-step explanation:

24x8=192

1/2 x24x8= 96

192-96=96

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I need help on this pleaseee
Leona [35]

Answer:

QUESTION:

I need help on this pleaseee...

ANSWER:

2 pairs of pants = $18

4 pairs of pants = $36

6 pairs of pants = $54

1 pair of pants = $6

Step-by-step explanation:

Hope that this helps you out! :)          

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2 years ago
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Un terreno rectangular cuyo ancho mide 20 mts menos que su largo. Su área del terreno es de 150 metros cuadrados. ¿Determina las
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2 years ago
H=65-1.4x what is the initial value?
wariber [46]
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8 0
3 years ago
17. Admission prices to Cinema I to see a movie are $9.50 for an adult and $6.50 for a child. The admission charge at Cinema II
maksim [4K]

Answer:

a. 9.5x + 6.5(x+c) < 8   when c>0

b. Must be one child more than the no. of adults.

Step-by-step explanation:

For Cinema 1:

for adult = $9.50

for child = $6.50

For Cinema 2:

Per person regardless of age = $8.00

First of all, we will find out the condition when per person rates in both cinema are equal.

Assume x = no. of adults

y = no. of children

Rate per person in Cinema I = Rate per person in Cinema II

(9.5x + 6.5y)/(x+y)   =   8

9.5x + 6.5y = 8(x+y)

9.5x + 6.5y = 8x + 8y

9.5x-8x = 8y-6.5y

=> x = y

So rates are equal when no. of adults equals no. of children

For Cinema I to have better rates, no. of children should be atleast 1 more than the no. of adult. In this way the rate per person of Cinema I will be less than 8

Hence we form an inequality when y = x+c and c > 0

9.5x + 6.5(x+c) < 8   when c>0

Hence there must be 1 more children than the no. of adults attending Cinema I for it to be a better deal.

4 0
3 years ago
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Approximately 4% of silicon wafers produced by a manufacturer have fewer than two large flaws. If Y, the number of flaws per waf
Doss [256]
Y ~ Po (λ)

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