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Tanya [424]
3 years ago
15

Tickets for the front section to a rock concert cost $25 each. The back section tickets sold for $15 each. If 400 tickets were s

old for a total revenue of $7,500, how many of each each type of ticket were sold? 1. Front – 145, Back – 255 2. Front – 140, Back – 260 3. Front – 155, Back – 245 4. Front – 150, Back – 250
Mathematics
1 answer:
frosja888 [35]3 years ago
8 0

Answer:

150 front tickets and 250 back tickets

Step-by-step explanation:

make 2 equations 25x + 15y = 7500 and x + y = 400 and do substitution on a graphing calculator or by your self.

let me know if this helps

You might be interested in
. Una persona desea realizar una pintura para su próxima exposición en donde esta tenga tres colores fluorescentes y dos colores
ycow [4]

Answer: B. 350.

Step-by-step explanation:

Given: Total number of fluorescent colors = 7

Total number of pastel colors = 5

Since , he want three fluorescent colors and two pastel colors.

Number of ways to select r things out of n things = ^nC_r=\dfrac{n!}{r!(n-r)!}

Number of ways to choose colors = ^7C_3\times\ ^5C_2

=\dfrac{7!}{3!4!}\times\dfrac{5!}{2!3!}\\\\=\dfrac{7\times6\times5\times4!}{3\times2\times1\times4!}\times\dfrac{5\times4\times3!}{2\times3!}\\\\=350

Hence, the correct option is B. 350.

3 0
2 years ago
Helppppp pleSeeeeeeeee
Zepler [3.9K]

Answer: C

Step-by-step explanation: Hope this help :D

3 0
3 years ago
2/3g + 1/2g = 14 multi step equations, i need help knowing the steps
MaRussiya [10]
Find the common denominator

It is 6

Multiply everything by 6

2/3*(6)= 4g

1/2 *(6)= 3g

14*(6)= 84

4g +3g = 84

7g= 84


g= 12
8 0
2 years ago
Someone help! It is urgent!!!
djverab [1.8K]

Answer:

a. f(-1)=12

b. f(2t)=16t²-6t+5

c. f(t-2)=4t²-19t+27

Step-by-step explanation:

For a, b, c, we are given an input. We plug that into f(x) to find our answers.

a.  f(-1)=12

f(-1)=4(-1)²-3(-1)+5

f(-1)=4+3+5

f(-1)=12

-------------------------------------------------------------------------------------------------------------

b. f(2t)=16t²-6t+5

f(2t)=4(2t)²-3(2t)+5

f(2t)=4(4t²)-6t+5

f(2t)=16t²-6t+5

-------------------------------------------------------------------------------------------------------------

c. f(t-2)=4t²-19t+27

f(t-2)=4(t-2)²-3(t-2)+5

f(t-2)=4(t²-4t+4)-3t+6+5

f(t-2)=4t²-16t+16-3t+6+5

f(t-2)=4t²-19t+27

7 0
2 years ago
The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and
natali 33 [55]

Answer:

a)\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03

p_v = P(\chi^2_{4} >17.03)=0.0019

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

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

b)

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Step-by-step explanation:

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".

Assume the following dataset:

Quality management        Excellent      Good     Fair    Total

Excellent                                40                35         25       100

Good                                      25                35         10         70

Fair                                         5                   10          15        30

Total                                       70                 80         50       200

Part a

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

H0: There is independence between the two categorical variables

H1: There is association between the two categorical variables

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

The statistic to check the hypothesis is given by:

\chi^2 = \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{70*100}{200}=35

E_{2} =\frac{80*100}{200}=40

E_{3} =\frac{50*100}{200}=25

E_{4} =\frac{70*70}{200}=24.5

E_{5} =\frac{80*70}{200}=28

E_{6} =\frac{50*70}{200}=17.5

E_{7} =\frac{70*30}{200}=10.5

E_{8} =\frac{80*30}{200}=12

E_{9} =\frac{50*30}{200}=7.5

And the expected values are given by:

Quality management        Excellent      Good     Fair       Total

Excellent                                35              40          25         100

Good                                      24.5           28          17.5        85

Fair                                         10.5            12           7.5         30

Total                                       70                 80         65        215

And now we can calculate the statistic:

\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03

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

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4} >17.03)=0.0019

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

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

Part b

We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:

Let's define some notation first.

E= Quality Management excellent     Ex=Reputation of company excellent

G= Quality Management good     Gx=Reputation of company good

F= Quality Management fait     Ex=Reputation of company fair

P(EΛ Ex) =40/215=0.186

P(GΛ Gx) =35/215=0.163

P(FΛ Fx) =15/215=0.0697

If we have dependence then the conditional probabilities would be higher values.

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

7 0
3 years ago
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