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3 years ago

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A certain company reported selling 5,000 toys during the month of January and expects sales to grow at a rate of 7.5% per month

until the end of that year. How many toys should the company expect to sell by the end of June of that year? 30,158 36,220 43,937 64,500

2 answers:

artcher [175]3 years ago

6 0

The formula is
Fv=p [(1+r)^(t)-1)÷r)
Fv ?
P 5000
R 0.075
T 6. June
Fv=5000 [(1+0.075)^(6)-1)÷0.075]
Fv=36220....answer

Alisiya [41]3 years ago

4 0

Answer:

Step-by-step explanation:

Alright, lets get started.

First month January, the toys are 5000.

In February, toys will be : $5000+\frac{5000*7.5}{100}$

In February, toys will be : 5375

In March, toys will be : $5375+\frac{5375*7.5}{100}$

In March, toys will be : 5778.125

In April, toys will be : $5778.125+\frac{5778.125*7.5}{100}$

In April, toys will be : 6211.484

In May, toys will be : $6211.484+\frac{6211.484*7.5}{100}$

In May, toys will be : 6677.345

In June, toys will be : $6677.345+\frac{6677.345*7.5}{100}$

In June, toys will be : 7178.146

Hence the total toys will be : $5000+5375+5778.125+6211.484+6677.345+7178.146$

Hence the total toys till June will be : 36220 : Answer

Hope it will help :)

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Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

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A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316$

$0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524$

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The margin of error for the proportion interval is given by this formula:

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And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32$

And rounded up we have that n=335

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And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599$

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