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mestny [16]
2 years ago
11

A retailer purchased a camera for rs.1560 and sold it at 5% loss find the loss amount​

Mathematics
1 answer:
zzz [600]2 years ago
3 0

Answer:

Loss is Rs. 78

Step-by-step explanation:

Solution:

Given,

  • Cost Price (CP)= Rs.1560
  • Loss Percent (L%) = 5%

Here,

L%= L/CP ×100%

or,. 5 =L/1560×100

or,. 5×1560/100 =L

or,. L= Rs.78

Therefore, Loss is Rs.78

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A waterslide is rented at $500 per day plus a charge per hour of use. The bounce house was used for 5 hours one day, and the tot
Mazyrski [523]

Answer: 500+5x=800

Step-by-step explanation:

Given: The rent of waterslide per day  = $500

Let x be the a charge per hour of use.

The number of hours bounce house was used  = 5 hours

Since, the total charge was $800.

Therefore, the equation can be used to find the charge x, in dollars, per hour of use is given by :-

Fixed charge+5 times x=Total charge

\Rightarrow500+5x=800

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mrs_skeptik [129]

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3 years ago
What is a non example of a rate?
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2 years ago
"The municipal transportation authority determined that 58% of all drivers were speeding along a busy street. In an attempt to r
vredina [299]

Answer:

a) X=77 drivers

b) Power of the test = 0.404

c) Increasing the sample size.

Step-by-step explanation:

This is a hypothesis test of proportions. As the claim is that the speed monitors were effective in reducing the speeding, this is a left-tail test.

For a left-tail test at a 5% significance level, we have a critical value of z that is zc=-1.645. This value is the limit of the rejection region. That means that if the test statistic z is smaller than zc=-1.645, the null hypothesis is rejected.

The proportion that would have a test statistic equal to this critical value can be expressed as:

p_c=\pi+z_c\cdot\sigma_p

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.58*0.42}{150}}\\\\\\ \sigma_p=\sqrt{0.001624}=0.04

Then, the proportion is:

p_c=\pi+z_c\cdot\sigma_p=0.58-1.645*0.04=0.58-0.0658=0.5142

This proportion, with a sample size of n=150, correspond to

x=n\cdot p=150\cdot0.5142=77.13\approx 77

The power of the test is the probability of correctly rejecting the null hypothesis.

The true proportion is 0.52, but we don't know at the time of the test, so the critical value to make a decision about rejecting the null hypothesis is still zc=-1.645 corresponding to a critical proportion of 0.51.

Then, we can say that the probability of rejecting the null hypothesis is still the probability of getting a sample of size n=150 with a proportion of 0.51 or smaller, but within a population with a proportion of 0.52.

The standard error has to be re-calculated for the new true proportion:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.52*0.48}{150}}\\\\\\ \sigma_p=\sqrt{0.001664}=0.041

Then, we calculate the z-value for this proportion with the true proportion:

z=\dfrac{p-\pi'}{\sigma_p}=\dfrac{0.51-0.52}{0.041}=\dfrac{-0.01}{0.041}=-0.244

The probability of getting a sample of size n=150 with a proportion of 0.51 or lower is:

P(p

Then, the power of the test is β=0.404.

The only variable left to change in the test in order to increase the power of the test is the sample size, as the significance level can not be changed (it is related to the probability of a Type I error).

It the sample size is increased, the standard error of the proprotion decreases. As the standard error tends to zero, the critical proportion tend to 0.58, as we can see in its equation:

\lim_{\sigma_p \to 0} p_c=\pi+ \lim_{\sigma_p \to 0}(z_c\cdot\sigma_p)=\pi=0.58

Then, if the critical proportion increases, the z-score increases, and also the probability of rejecting the null hypothesis.

5 0
3 years ago
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