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Kaylis [27]
2 years ago
14

20 pts Help me please

Mathematics
1 answer:
Viktor [21]2 years ago
8 0

Answer:

4<em>s</em> + 10<em>p</em> <= 100

<em>p</em> >= 5

Step-by-step explanation:

So it's important to remember here that this isn't one linear inequality to represent everything, it's a system. One of the inequalities that are part of the answer won't answer the entire question by itself, so don't worry about that. It might sound like tedious work, but you can break it down pretty easily into logic.

So the soil costs Marsha $4 per bag, and the plants cost her $10 each. She can't spend more than $100 dollars.

Let <em>p</em> = number of plants and let <em>s</em> = number of bags of soil

If each soil bag costs her $4, then 4<em>s</em> is the amount she spends on soil (4 [which is the cost of each individual soil bag] x number of bags = total cost of soil bags).

If each plant costs her $10, then 10<em>p</em> is the amount she spends on plants (same logic).

Because she can't spend more than $100 on both soil and plants together,

4<em>s</em> + 10<em>p</em> has to be less than or equal to 100.

4<em>s</em> + 10<em>p</em> <= 100

Since she wants to buy at least 5 plants, <em>p</em> has to be greater than or equal to 5.

<em>p</em> >= 5

And there you have it! Your system of inequalities is:

4<em>s</em> + 10<em>p</em> <= 100

<em>p</em> >= 5

Pretty simple c:

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Answer:

10 1/4 - 2 2/4 = 37/4

Step-by-step explanation:

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  3. (41/4) - 1 = 37/4

That answer!

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The question is an illustration of equivalent ratios.

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So, we have:

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So, we have:

Red = 3oz \times 1.5 = 4.5oz

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brainly.com/question/18441891

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Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
My name is Ann [436]

Answer:

a. The value of alpha is 3.014 and the value of beta is 12.442

b. The probability that data transfer time exceeds 50ms is 0.238

c. The probability that data transfer time is between 50 and 75 ms is 0.176

Step-by-step explanation:

a. According to the given data we have that the mean and standard deviation of the random variable X are 37.5 ms and 21.6.

Therefore, E(X)=37.5 and V(X)=(21.6)∧2  

To calculate alpha we would have to use the following formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To calculate beta we would have to use the following formula:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. E(X)=37.5 and V(X)=(21.6)∧2  

Therefore, P(X>50)=1−P(X≤50)

Hence, To calculate the probability that data transfer time exceeds 50ms we use the following formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The probability that data transfer time exceeds 50ms is 0.238

c. E(X)=37.5 and V(X)=(21.6)∧2  

​Therefore, P(50<X<75)=P(X<75)−P(X<50)  

Hence, To calculate the probability that data transfer time is between 50 and 75 ms we use the following formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time is between 50 and 75 ms is 0.176

6 0
3 years ago
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