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olga_2 [115]
3 years ago
5

Grant bought more than 120 flower bulbs to plant in his garden. He bought 2 bags of tulip bulbs with 30 bulbs per bag. He also b

ought 5 bags of daffodil bulbs. What is the least number of daffodil bulbs that could be in each bag?
Mathematics
1 answer:
podryga [215]3 years ago
8 0

Answer:

The minimum number of daffodils in each bag have to be 12.

Step-by-step explanation:

Given:

Grant bought more than 120 flower bulbs to plant in his garden.

He bought 2 bags of tulip bulbs with 30 bulbs per bag.

He also bought 5 bags of daffodil bulbs.

To find:

Least number of daffodil bulbs that could be in each bag = ?

Solution:

So, we know that Grant has more than 120 total flowers in his garden which means he has a minimum of 120 flowers in his garden. So, for calculation we will consider the total number of flowers in his garden as 120.

But we need to find the minimum number of daffodils in each bag so that all the flowers add up to 120.

Let the number of daffodils in each bag be ‘x’.

So, from the given information we can form the following expression.

Tulips + daffodils =120

The number of tulip flowers are:

=2×30

=60

Therefore, total number of flowers:

60 + 5(x)=120

5(x)=120-60

5(x)=60

x=12

Hence the minimum number of daffodils in each bag have to be 12.

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g: Consider the following game. A coin is flipped. If it lands tails we stop playing and winnothing. If it lands heads we spin a
galben [10]

Answer:

E[W] = $25 (assuming the currency is in dollars)

Var(W) = 1041.67

Step-by-step explanation:

Probability of winning first starts with the coin toss.

For a win, the coin needs to land on heads.

Probability of that = 1/2 = 0.5

Then probability of winning any amount = 1/100 = 0.01

Total probability of winning any amount = 0.5 × 0.01 = 0.005

But expected value is given by

E(X) = Σ xᵢpᵢ

where xᵢ is each amount that could be won

pᵢ is the probability of each amount to be won and it is the same for all the possible winnings = 0.005

So,

E(W) = Σ 0.005 xᵢ

Summing from 0 to 100 (0 indicating getting a tail from the coin toss). This could be done with dome faster with an integral sign

E(W) = ∫ 0.005 x dx

Integrating from 0 to 100

E(W) = [0.005 x²/2]¹⁰⁰₀

E(W) = [0.0025 x²]¹⁰⁰₀ = 0.0025(100² - 0²) = 0.0025 × 10000 = $25

Variance is given by

Variance = Var(X) = Σxᵢ²pᵢ − μ²

μ = expected value

We calculate the expression, Σxᵢ²pᵢ which is another sum from 0 to 100

Σxᵢ²pᵢ = Σ 0.005xᵢ²

Σ 0.005 xᵢ² = ∫ 0.005 x² dx

Integrating from 0 to 100

∫ 0.005 x² dx = [0.005 x³/3]¹⁰⁰₀ = [0.1667x³]¹⁰⁰₀ = 0.1667(100³ - 0³) = 1666.67

Var(W) = 1666.67 - 25² = 1666.67 - 625 = 1041.67.

5 0
3 years ago
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