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2 years ago

MR Arsalan bought a plot of land for rs 108 000 and sold it for a profit of 20% at what price did he sold at

Mathematics

1 answer:

disa [49]2 years ago

6 0

Answer:

The plot of land was sold for rs 129,600

Step-by-step explanation:

The cost of the plot of land = rs 108,000

Profit calculation ( 20% x 108,000) = rs 21,600

Price land was sold = (108,000 + 21,600) = 129,600

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In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. In the survey,

balandron [24]

Answer:

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

And for this case the confidence interval is given by: (0.275, 0.305)

We can estimate the proportion difference as:

$\hat p_D = \frac{0.275+0.305}{2}=0.29$

And the margin of error would be:

$ME=0.305-0.29=0.015$

So then for this case the possibl two options are:

We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%.

We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.

Step-by-step explanation:

Previous concepts

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.

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".

$p_A$ represent the real population proportion of Latinos who view global warming as a serious problem

$\hat p_A=0.75$ represent the estimated proportion Latinos who view global warming as a serious problem

$n_A$ is the sample size required of Latinos who view global warming as a serious problem

$p_B$ represent the real population proportion of white who view global warming as a serious problem

$\hat p_B =0.46$ represent the estimated proportion of whitewho view global warming as a serious problem

$n_B$ is the sample size required of white

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

And for this case the confidence interval is given by: (0.275, 0.305)

We can estimate the proportion difference as:

$\hat p_D = \frac{0.275+0.305}{2}=0.29$

And the margin of error would be:

$ME=0.305-0.29=0.015$

So then for this case the possibl two options are:

We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%.

We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.

4 0

2 years ago

What is the quotient of 5472 and 81?

Akimi4 [234]

Answer:

Step-by-step explanation:

Division:-

$\sf\Large\qquad\quad67\\ \begin{array}{cc} \cline{2 - 2}\sf 81 )&\sf \ 5472\\&\sf -487 \downarrow\\ \cline{2-2}& \sf \ \ \ \ 602\\ &\sf \ - 567 \\ \cline{2-2} & \sf \ \035 \\ \cline{2-2} \end{array}\\\\\\ Divisor \rightarrow 81 \\ \\ Quotient \rightarrow 67\\\\Remainder \rightarrow 35$

5 0

1 year ago

A game has 15 balls for each of the letters B, I, N, G, and O. The table shows the results of drawing balls 1,250 times.

Mariana [72]

Theoretically, each letter should have the same probability of occurring since there are 15 of each. There are 5 letters that can be drawn, so there is a total of 75 balls, and each letter has a probability $\dfrac{15}{75}=\dfrac15$ of being drawn.

This means one would expect a theoretical frequency of $\dfrac{1250}5=250$ , i.e. any given letter should get drawn 250 times, which means the answer is D.

4 0

2 years ago

In a normally distributed data set a mean of 55 where 95% of the data fall between 47.4 and 62.6, what would be the standard dev

tresset_1 [31]

Answer:

The standard deviation of that data set is 3.8

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 55

95% of the data fall between 47.4 and 62.6. This means that 47.4 is 2 standard deviations below the mean and 62.6 is two standard deviations above the mean.

Using one of these points.

55 + 2sd = 62.6

2sd = 7.6

sd = 7.6/2

sd = 3.8

The standard deviation of that data set is 3.8

6 0

2 years ago

A taxi travels 2 hours to make a 220 km trip. How fast does it travel?<br>

sertanlavr [38]

He would be driving approximately 110 km an hour

7 0

1 year ago

Read 2 more answers