If 4, 6 and 8 and 14, 21, and x are the lengths of the corresponding sides of two similar triangles, what is the value of x

The cost of, y, of a large pizza with x topping from pats pizzeria can be modeled by a linear function. A large pizza with no to

Firdavs [7]

Answer:

The cost would be 22.75

Step-by-step explanation:

To find the cost of each topping, we simply take the excess cost and divide it by the number of toppings. To find the excess cost, take the new cost and subtract the original cost.

17.50 - 14.00 = 3.50

Now divide that by the amount of toppings.

3.50/2 = 1.75

Now we can multiply the new number of toppings by that amount and add the cost of the pizza.

5(1.75) + 14.00

8.75 + 14.00

22.75

3 0

4 years ago

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Will mixes 1/4 gallon of milk and 1/2 gallon of chocolate syrup to make chocolate milk. Each serving is 1/16 gallon. How many se

Dennis_Churaev [7]

Answer:

12 servings!

Step-by-step explanation:

1/4 equals 0.25 gallons, 1/2 gallons is 0.5, so Will will have a mix of 0.75 gallons. If each serving is 1/16 and 1/16 equals 0.0625, divide 0.75 by 0.0625 = 12 servings.

6 0

3 years ago

a sample of unknown liquid has a volume of 12.0ml and a mass off 6g, what is its density? and how did you do the problem?

maw [93]

Answer:

Step-by-step explanation:

The equation for solving for density is:

Density = mass / volume

Then plug in the values:

Density = 6 g / 12 mL

Divide and don't forget the units!

Density = 0.5 g/mL

5 0

4 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago

Please help! time sensitive.

Ronch [10]

Answer:

true

Step-

when you divide fractions you have to keep change flip

7 0

3 years ago

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