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Vesna [10]
3 years ago
7

2 At Bea's Pet Shop, the number of dogs, d, is initially five less than twice the number of cats, c. If she decides to add three

more of each, the ratio
of cats to dogs will be – Write an equation or system of equations that can be used to find the number of cats and dogs Bea has in her pet shop. Could Bea's Pet Shop initially have 15 cats and 20 dogs? Explain your reasoning. Determine algebraically the number of cats and the number of dogs Bea initially had in her pet shop.​
Mathematics
1 answer:
forsale [732]3 years ago
7 0

Let D be dogs and C be cats

<em>dogs, d, is initially five less than twice the number of cats, c</em>

D + 5 = 2C

<em>If she decides to add three more of each, the ratio  of cats to dogs will be</em>

D + 8 = 2C + 3

<em>Could Bea's Pet Shop initially have 15 cats and 20 dogs?</em>

Simply plug in the numbers

20 + 5 = 2(15)

This is clearly not true: 25 does not equal 30

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Answer:

i) 2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959  

2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041  

So then the 90% confidence interval is given by (1.959, 2.041)

ii) Figure attached

iii) ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}   (1)

And for a 90% of confidence the significance is given by \alpha=1-0.9=0.1, and \frac{\alpha}{2}=0.05. Since we know the population standard deviation we can calculate the critical value z_{0.05}= \pm 1.64

We know the folllowing data:

\bar X = 2 represent the sample mean

\sigma = 0.1 represent the population deviation

n =16 represent the sample size

Part i)

If we replace the values given into formula (1) we got:

2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959  

2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041  

So then the 90% confidence interval is given by (1.959, 2.041)

Part ii)

Figure attached. We have the illustration for the confidence interval obtained.

Part iii)

The margin of error is given by:

ME=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}

And if we replace we got:

ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041

3 0
4 years ago
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