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irina1246 [14]
3 years ago
10

Answer quick please.

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
4 0

Answer

The Answer is A C D

Step-by-step explanation:

You might be interested in
Atlantic auditorium has 850 seats. Tickets were sold for 816 of the seats. For what percent of the seats were the tickets sold?
Mariana [72]

Answer:

<u>96 percent</u> of the seats were the tickets sold.

Step-by-step explanation:

Given:

Atlantic auditorium has 850 seats. Tickets were sold for 816 of the seats.

Now, to find percent of the seats of the tickets sold.

Total number of seats = 850.

Tickets sold for the total seats = 816.

So, to find the percent of the seats of the tickets sold:

\frac{816}{850}\times 100

=0.96\times 100

=96\%.

Therefore, 96 percent of the seats were the tickets sold.

5 0
3 years ago
Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa
xxMikexx [17]

Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent 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

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

_____________________________________

Total                                     189    36   225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

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{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

And now we can calculate the statistic:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0
4 years ago
Lena needs $50 to buy some notebooks. She had saved $2 and plans to work as a babysitter to earn $8 per hour. Which inequality s
fgiga [73]
<span> 2 + 8n ≥ 50, so n ≥ 6</span>
4 0
3 years ago
Nia has $19.50 to ride the subway around New York. It will cost her $0.75 every time she rides. Identify the dependent variable
ra1l [238]

Answer:

The correct answer is B. The number of rides and the total cost are both dependent variables, as both depend on the amount of money that Nia has.

Step-by-step explanation:

The total cost is dependent because it relies on how many rides Nia takes, and the $19.50 limit.

The number of rides is dependent on the limit of $19.50 because the $19.50 is the limit and the total amount of rides is locked by $19.50

I'm not sure, though.

5 0
3 years ago
Read 2 more answers
Please answer There is a bag filled with 3 blue and 5 red marbles. A marble is taken at random from the bag, the colour is noted
Misha Larkins [42]

Answer:

\frac{15}{28}  is the required probability.

Step-by-step explanation:

Total number of Marbles = Blue + Red = 3 + 5 = 8

Probability of getting blue = \frac{3}{8}

Probability of not getting a blue =\frac{5}{8}

To get exactly one blue in two draws, we either get a blue, not blue, or a not blue, blue.

<u>First Draw Blue, Second Draw Not Blue:</u>

1st Draw: P(Blue) = \frac{3}{8}

2nd Draw: P(Not\:Blue)=\frac{5}{7}  (since we did not replace the first marble)

To get the probability of the event, since each draw is independent, we multiply both probabilities.

P(Event)=\frac{3}{8}\cdot \frac{5}{7}=\frac{15}{56}

<u>First Draw Not Blue, Second Draw Not Blue:</u>

1st Draw: P(Not\:Blue)=\frac{5}{8}

2nd Draw: P(Not\:Blue)=\frac{3}{7}  (since we did not replace the first marble)

To get the probability of the event, since each draw is independent, we multiply both probabilities.

P(Event)=\frac{5}{8}\cdot \frac{3}{7}=\frac{15}{56}

To get the probability of exactly one blue, we add both of the events:

\frac{15}{56}+\frac{15}{56}=\frac{15}{28}

4 0
3 years ago
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