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USPshnik [31]
3 years ago
8

At a garden shop, a customer buys 5 geraniums and 4 lilies for $48. Another customer pays $58 for 4 geraniums and 6 lilies. Writ

e and solve system of equations to find the cost of one geranium and one lily
Mathematics
1 answer:
masha68 [24]3 years ago
7 0

Answer:

A lily costs $7 and a geranium $4.

Step-by-step explanation:

From the question, we can write two equations. let the number of lilies be l and the number of geraniums be g, then:

5 g + 4 l = 48

4 g + 6 l = 58

Multiply the first equation by 4 and the second by 5, the number of lilies in the other gives:

20 g + 16 l = 192

20 g + 30 l = 290

Subtract the first equation from the second gives:

14 l = 98  which dividing by 14 gives  l = 7

Substituting the value  l = 7  in the first equation gives:

5 g + 28 = 48

Subtract 20 from both sides gives:

5 g = 20  divide by 5 gives  g = 4

So, a lily costs $7 and a geranium costs $4.

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According to a recent​ publication, the mean price of new mobile homes is ​$63 comma 800. Assume a standard deviation of ​$7900.
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Answer:

a. For n=25, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,580​, respectively.

b. For n=50, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,117​, respectively.

Step-by-step explanation:

In this case, for each sample size, we have a sampling distribution (a distribution for the population of sample means), with the following parameters:

\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{n}

For n=25 we have:

\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{25}=7,900/5=1,580

The spread of the sampling distribution is always smaller than the population spread of the individuals. The spread is smaller as the sample size increase.

This has the implication that is expected to have more precision in the estimation of the population mean when we use bigger samples than smaller ones.

If n=50, we have:

\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{50}=7,900/7.07=1,117

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