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4 years ago

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Jake, Sara, and Cole are selling tickets for the school drama show. Sara sells half as many tickets as Jake. Cole sells 6 more t

han twice as much as Jake. They sold a total of 538 tickets together.
a) Write an equation to represent the situation.
b) Solve to find how many tickets Sara sold.

2 answers:

3241004551 [841]4 years ago

7 0

We let x be the number of tickets that Sara and Jake sold individually. With this representation, the number of tickets sold by Cole is equal to 2x+6.

(a) x + x + (2x + 6) = 538
The value of x from the generated equation is 133.

(b) Sara sold a total of 133 tickets.

andrew11 [14]4 years ago

4 0

Answer:

Let x be the number of tickets Jake sold,

Since, Sara sells half as many tickets as Jake.

⇒ The ticket sold by Sara = $\frac{x}{2}$

Also, Cole sells 6 more than twice as much as Jake.

⇒ The ticket sold by Cole = 2x + 6,

Thus, the ticket sold by all three = $x+\frac{x}{2}+2x+6$

a) According to the question,

$x+\frac{x}{2}+2x+6=538$

b) By solving this equation,

$\frac{2x+x+4x+12}{2}=538$

$\frac{7x+12}{2}=538$

$7x+12=1076$

$7x=1064$

$\implies x=\frac{1064}{7}=152$

Thus, the number of tickets Sara has = $\frac{x}{2}=\frac{152}{2}=76$

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A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand

sesenic [268]

Answer:

The 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

Step-by-step explanation:

Previous concepts

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".

$\bar X_1 =80$ represent the sample mean 1

$\bar X_2 =75$ represent the sample mean 2

n1=25 represent the sample 1 size

n2=36 represent the sample 2 size

$\sigma_1 =5$ sample standard deviation for sample 1

$\sigma_2 =3$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =80-75=5$

Since the Confidence is 0.94 or 94%, the value of $\alpha=0.06$ and $\alpha/2 =0.03$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that $z_{\alpha/2}=1.88$

The standard error is given by the following formula:

$SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118$

Confidence interval

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

$5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898$

$5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102$

So on this case the 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

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