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melisa1 [442]
3 years ago
15

Selene wants to invite some friends to go with her to see the El Control concert. She will purchase tickets online for $42.50 ea

ch plus a $5 handling fee. If Selene has $250, how many tickets can she buy? Write and solve an inequality to model the situation.
A
42.50x+5≤250

5 tickets

B
42.5x+5≥250

6 tickets

C
42.50x+5≥250

5 tickets

D
42.50x+5≤250

6 tickets
Mathematics
1 answer:
nikdorinn [45]3 years ago
6 0
The price of each ticket is $42.50. So if Selene buys x tickets, the cost of x tickets will be 42.50x.

She has to pay $5 handling fee as well for purchasing the tickets online. So her total cost of the tickets will be:

Total Cost = 42.50x + 5

Selene got $250. This means she can spend a maximum of $250 on the tickets. Maximum of $250 means less than or equal to $250.

So, we can write the spending on the tickets in the form of following in equality:

42.50x + 5 ≤ 250

42.50x ≤ 245

x ≤ 5.76

This shows, Selene can buy 5.76 tickets. Since the number of tickets has to be a whole number we look for the nearest possible number. She cannot buy 6 tickets, as the cost will exceed $250. So the maximum tickets she can buy are 5.

Therefore, the correct answer is option A
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tester [92]

Answer:

StartFraction 24 Over 65 EndFraction

Step-by-step explanation:

Total number of students = 26

Number of boys = 10

Number of girls = 26-10

=16

Eduardo has to pull two names out of the hat without replacing them.

First name

Probability= Favourable outcome/Total outcome

Probability of girls=16

Total probability=26

Eduardo has to pull two names out of the hat without replacing them.

Probability= Favourable outcome/Total outcome

=16/26

=8/13

For the second name:

Without replacement of the first hat

Probability of girls=16-1=15

Total probability=26-1=25

Probability= Favourable outcome/Total outcome

=15/25

=3/5

Probability of both of Eduardo's partner for the group project will be girls=8/13*3/5

=24/65

StartFraction 24 Over 65 EndFraction

4 0
3 years ago
The following data lists the ages of a random selection of actresses when they won an award in the category of Best​ Actress, al
Valentin [98]

Answer:

a) p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation  

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \geq 0

Alternative hypothesis: \mu_y -\mu_x

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679

The next step is calculate the degrees of freedom given by:

df=n-1=10-1=9

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The confidence interval for the mean is given by the following formula:  

\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that t_{\alpha/2}=3.25  

Now we have everything in order to replace into formula (1):  

-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469  

-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069  

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

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Step-by-step explanation:

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We need to have the same units

convert yd to ft

1 yd = 3ft

Multiply each by 70

70 yds = 210 ft

A = 210 *20

A = 4200 ft^2



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