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zhuklara [117]
3 years ago
6

Tyler ordered a pizza to eat while he watches a movie. Before the movie began, he ate 1/4 of the pizza. During the movie, he ate

another 3/8 Before storing the remaining pizza, he ate a final 1/16 of the pizza. How much leftover pizza does Tyler have.
Mathematics
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

Tyler has \frac{5}{16} Left over Pizza.

Step-by-step explanation:

Given:

Total pizza = 1

Pizza ate at First time = \frac{1}{4}

Pizza ate at Second time = \frac{3}{8}

Pizza ate at Final time = \frac{1}{16}

We need to find find the amount of pizza left.

Now to find the Amount of Pizza left is equal to Total Pizza minus Pizza ate at First time minus Pizza ate at second time minus Pizza ate at Final time.

Framing in the equation form we get;

Amount of Pizza left = 1- \frac{1}{4} -\frac{3}{8} -\frac{1}{16}

Now Taking the LCM we get;

Amount of Pizza left = \frac{1\times16}{16}- \frac{1\times4}{4\times 4} -\frac{3\times2}{8\times2} -\frac{1\times1}{16\times1}= \frac{16}{16}- \frac{4}{16} -\frac{6}{16} -\frac{1}{16}= \frac{16-4-6-1}{16}= \frac{5}{16}

Hence Tyler has \frac{5}{16} Left over Pizza.

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If the product of roots of the equation given below is 4, then find the value of m.
Vikentia [17]

Answer:

m=-2

Step-by-step explanation:

As the product of the roots of a quadratic equation is c/a in ax^2+bx+c=0

here a=2, b=+8, c=-m^3

Given c/a=4

-m^3/2=4

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m=-2.

8 0
3 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
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What is the area of the figure?
Bad White [126]
20m ^2

Explanation:
The top part be be cut into 2 2•2 triangles. A = 4 for both of them
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4 + 16 = 20

Hope this helps!
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3 years ago
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1 x 10^10 or 10000000000
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