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3 years ago

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A charity is holding a benefit at an Italian restaurant. The owner of the restaurant offers to donate $200 in addition to $4 for

each person that
attends. Write an equation where d is the total donation and p in the number of people that attend.

1 answer:

iren [92.7K]3 years ago

5 0

Answer:

200 divided 4

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The slope intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

We have the equation of a line in the standard form

$Ax+By=C$

Convert to the slope-intercept form:

$4x+2y=12$ <em>subtract 4x from both sides</em>

$2y=-4x+12$ <em>divide both sides by 2</em>

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Other method:

If we have the equation of a line in standard form Ax + By = C

or in general form Ax + By + C = 0, then the formula of a slope is

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Step-by-step explanation: :)

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Read 2 more answers

Researchers obtained a sample of n = 10 children who attended daycare before starting school. The children are given a standardi

attashe74 [19]

Answer:

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$

$p_v =2*P(t_{9}>4.093)=0.00084$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is significantly different from 50.

Step-by-step explanation:

Data given and notation

53, 57, 61, 49, 52, 56, 58, 62, 51, 56

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X))^2}{n-1}}$

And we got:

$\bar X=55.5$ represent the sample mean

$s=4.249$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =50/tex] represent the value that we want to test [tex]\alpha=0.05$ represent the significance level for the hypothesis test (ASSUMED).

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal or not to 50 :

Null hypothesis: $\mu = 50$

Alternative hypothesis: $\mu \neq 50$

Since we don't know the population deviation, 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".

Calculate the statistic

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

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=10-1=9$

Since is a two-sided tailed test the p value would given by:

$p_v =2*P(t_{9}>4.093)=0.00084$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is significantly different from 50.

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