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AlexFokin [52]
3 years ago
11

The variable x represents the number of angelfish Carlos bought and the variable y represents the number of parrotfish he bought

.
Carlos bought 405 tropical fish for a museum display. He bought 8 times as many parrotfish as angelfish.

How many of each type of fish did he buy
Mathematics
1 answer:
expeople1 [14]3 years ago
6 0
If you would like to know how many of each type of fish did Carlos buy, you can calculate this using the following steps:

x ... the number of angelfish
y ... the number of parrotfish
405 tropical fish bought
8 times as many parrotfish as angelfish ... 8 * y = x

405 = x + y
405 = 8 * y + y
405 = 9 * y
y = 405 / 9
y = 45 parrotfish

x = 8 * y = 8 * 45 = 360 angelfish

The correct result would be 45 parrotfish and 360 angelfish.
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A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e
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a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem  guarantees that \bar x is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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 95% of confidence the significance is given by \alpha=1-0.95=0.05, and \frac{\alpha}{2}=0.025. Since we know the population standard deviation we can calculate the critical value z_{0.025}= \pm 1.96

Part a

n=40,\bar X=42,\sigma=50

If we use the formula (1) and we replace the values we got:

42 - 1.96 \frac{50}{\sqrt{40}}=26.50  

42 + 1.96 \frac{50}{\sqrt{40}}=57.50  

The 95% confidence interval is given by (26.50;57.50)

Part b

n=300,\bar X=123,\sigma=50

If we use the formula (1) and we replace the values we got:

123 - 1.96 \frac{50}{\sqrt{300}}=117.34  

123 + 1.96 \frac{50}{\sqrt{300}}=128.66  

The 95% confidence interval is given by (117.34;128.66)

Part c

n=155,\bar X=20,\sigma=50

If we use the formula (1) and we replace the values we got:

20 - 1.96 \frac{50}{\sqrt{155}}=12.13  

20 + 1.96 \frac{50}{\sqrt{155}}=27.87  

The 95% confidence interval is given by (12.13;27.87)

Part d

n=155,\bar X=3.14,\sigma=50

If we use the formula (1) and we replace the values we got:

3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73  

3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01  

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem  guarantees that \bar x is approximately normal, so the confidence intervals are valid

8 0
3 years ago
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