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Marrrta [24]
3 years ago
8

Keta sold tickets for her school play at $6 for students, and $9 for adults. She sold 15 adult tickets and made a sales total of

$315. How many student tickets were sold 48 points
Mathematics
2 answers:
frosja888 [35]3 years ago
6 0

Answer:I want to help you, I just don’t know how to because there is not enough information.

Step-by-step explanation

ss7ja [257]3 years ago
5 0

Answer:

Let x be the number of tickets sold to adults and y the number sold to students:

x + y = 315. But we know that y was double x, then :

y = 2x

Plug the value of y (in the 1st equation) by 2x:

 x+ 2x = 315

3x = 315

and x = 105 (which is the number of tickets sold to adults

Read more on Brainly.com - brainly.com/question/4429543#readmore

Step-by-step explanation:

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(2x+5)-3

Step-by-step explanation:

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3 years ago
Reflect AQ'R'S' if Q'(1, -3), R'(-2,-6),<br> and S'(-1,-1) over the x-axis.
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Step-by-step explanation:

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2 years ago
Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 a
krok68 [10]

Answer:

Jessie scored higher than Reagan.

Step-by-step explanation:

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. <u>Finding z-score for Reagan;</u>

Let X = distribution of SAT scores

SO, X ~ Normal(\mu=1000, \sigma^{2}=100^{2})

The z-score probability distribution for the normal distribution is given by;

                                    Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 1000

            \sigma = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

       z-score  =  \frac{1140-1000}{100}  =  1.4

2. <u>Finding z-score for Jessie;</u>

Let X = distribution of ACT scores

SO, X ~ Normal(\mu=17, \sigma^{2}=5^{2})

The z-score probability distribution for the normal distribution is given by;

                                    Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 17

            \sigma = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

       z-score  =  \frac{30-17}{5}  =  2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.

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Answer:

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