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kondaur [170]
3 years ago
8

A regular meal at a restaurant costs $6. A deluxe meal at the restaurant costs $3 than a regular meal. A group visits the restau

rant and orders regular and deluxe meals totaling $249
Mathematics
1 answer:
wolverine [178]3 years ago
3 0
Their could be 14 people who order the deluxe and 18 people who get regular meals
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Show ur work and check it for both
Oksi-84 [34.3K]

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7) is 26 8) is 20

30-4=26

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3 years ago
Quadratic applications??
aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

2 l + 2w = 34 ⇔ l + w = 17 ⇒ l = 17 - w ..... <em>(1)</em>

l w = 66 ..... <em>(2)</em>

<em>(1)</em> ----> <em>(2)</em>

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5 0
3 years ago
PLEASE help me with this problem and explain the steps you used.
Bad White [126]
I hope this helps you




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7 0
4 years ago
an Acrison dry additive feeder is delivering a material at a rate of 28 cubic inches per minute. If the material has a bulk dens
LiRa [457]

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oof

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4 0
3 years ago
The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams and standard
yuradex [85]

Answer:

Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)

Step-by-step explanation:

Given:

Weights of Broilers are normally distributed.

Mean = 1387 g

Standard Deviation = 161 g

To find: Probability that a randomly selected broiler weighs more than 1454 g.

we have ,

Mean,\,\mu=1387

Standard\,deviation,\,\sigma=161

X = 1454

We use z-score to find this probability.

we know that

z=\frac{X-\mu}{\sigma}

z=\frac{1454-1387}{161}=0.416=0.42

P( z = 0.42 ) = 0.6628   (from z-score table)

Thus, P( X ≥ 1454 ) = P( z ≥ 0.42 ) = 1 - 0.6628 =  0.3372

Therefore, Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)

8 0
3 years ago
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