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dangina [55]
3 years ago
7

30 children are going on a school trip

Mathematics
1 answer:
Kruka [31]3 years ago
8 0
30 × 5 = 150

150 - 110 = 40

40 ÷ 2 = 20

So 20 kids brought their own lunch

Hope this helped. Have a great day!

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3+2x=17 I’m taking a test
mihalych1998 [28]

Answer: x=7

Step-by-step explanation:

2x=17-3

2x=14

X=7

8 0
3 years ago
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Your surf shop sells two types of surfboards. The first type A, costs $272 and you make a profit of $29 on each one. The second
brilliants [131]
Let the number of type A surfboards to be ordered be x and the number of type B surfboards be y, then we have

Minimize: C = 272x + 136y

subject to: 29x + 17y ≥ 1210
                 x + y ≤ 50
                 x, y ≥ 1

From the graph of the constraints, we have that the corner points are:

(20, 30), (41.138, 1) and (49, 1)

Applying the corner poits to the objective function, we have

For (20, 30): C = 272(20) + 136(30) = 5440 + 4080 = $9,520
For (41.138, 1): C = 272(41.138) + 136 = 11189.54 + 136 = $11,325.54
For (49, 1): C = 272(49) + 136 = 13328 + 136 = $13,464

Therefore, for minimum cost, 20 type A surfboards and 30 type B surfboards should be ordered.
6 0
3 years ago
Order these decimals form greatest to least 3.6; 0.36; 36; 0.036
aliina [53]
36, 3.6, 0.36, 0.036 is the order
4 0
3 years ago
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One household is to be selected at random from a town. ​ ​The probability that ​the household has a cat is 0.20.2 . ​ ​The proba
dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a ​cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

A = a + (A \cap B)

In which a is the probability that a household has a cat but not a dog and A \cap B is the probability that a household has both a cat and a dog.

By the same logic, we have that:

B = b + (A \cap B)

The probability that the household has a cat or a dog is 0.5

a + b + (A \cap B) = 0.5

The probability that the household has a dog ​is 0.4

B = 0.4

B = b + (A \cap B)

b = 0.4 - (A \cap B)

The probability that ​the household has a cat is 0.2.

A = 0.2

A = a + (A \cap B)

a = 0.2 - (A \cap B)

So

a + b + (A \cap B) = 0.5

0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5

A \cap B = 0.1

What is ​the probability that the household has a dog, given that the household has a ​cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5

There is a 50% probability that the household has a dog, given that the household has a ​cat.

4 0
3 years ago
Helen earns $12 each weekend by babysitting her brother.After the third weekend,Helen buys a new cd for $12.48.How much money do
GalinKa [24]
She would have $23.52 because 12 times 3 is 36 and 36-12.48=23.52
6 0
3 years ago
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