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photoshop1234 [79]

2 years ago

8

A man bought 550 oranges at 60 paisa each and 240 oranges at 50 paisa each. If he sold them at the rate of 10 for Rs 9. what was

his profit percentage? Please show step by step .

1 answer:

Hunter-Best [27]2 years ago

6 0

Answer:

hence the profit percent is 58%.

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Answer:

a) Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

b) $X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c) $t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

$p_v =P(t_{(49)}>2.81)=0.0035$

d) If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X=32.12$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =30.2$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

$X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c. Perform the test and calculate P-value

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(49)}>2.81)=0.0035$

d. State your conclusion.

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

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3 years ago

A dump truck brought 1⁄3 of a ton of rock on the first trip, 1⁄2 of a ton on the second trip, but on the third trip, had to take

Allisa [31]

Answer:

1/30 of a ton

Step-by-step explanation:

Write this as an expression:

1/3 of a ton and 1/2 of a ton were taken to the site.

4/5 of a ton were remove from the site.

Amount of a ton at the site = $\frac{1}{3}+ \frac{1}{2}- \frac{4}{5}$

Solve the equation by finding <u>common denominators</u> (when the bottom numbers are the same).

$(\frac{1}{3}+ \frac{1}{2})- \frac{4}{5}$ Focus on the first part. The least common multiple (LCM) of 3 and 2 is "6", which will become the denominator.

For <u>1/3 to become ?/6</u>, multiply top and bottom by 2.

For <u>1/2 to become ?/6</u>, multiply top and bottom by 3.

$(\frac{2}{6}+ \frac{3}{6})- \frac{4}{5}$

Now we can add the numerators together over the same denominator.

$\frac{5}{6}- \frac{4}{5}$

Do the same thing as before and change the denominators. The LCM of 6 and 5 is "30".

For <u>5/6 to become ?/30</u>, multiply top and bottom by 5.

For <u>4/5 to become ?/30</u>, multiply top and bottom by 6.

$\frac{25}{30}- \frac{24}{30}$

Subtract the numerators:

$\frac{1}{30}$ Answer in tons

Therefore there was 1/30 of a ton of rock left at the construction site.

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