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Elenna [48]
3 years ago
11

A soccer team sold raffle tickets to raise money for the upcoming season. They sold three different types of tickets: premium fo

r $6, deluxe for $4, and regular for $2. The total number of tickets sold was 273, and the total amount of money from raffle tickets was $836. If 118 more regular tickets were sold than deluxe tickets, how many premium tickets were sold?
Mathematics
2 answers:
puteri [66]3 years ago
8 0

Answer:

Step-by-step explanation

We get three linear equations from the information given, where

p= number of premium tickets

d = number of deluxe tickets

r = number of regular tickets:

\left \{ {{p+d+r=273} \atop \\{6p+4d+2r=836} \right.

and the applying third r=118+d, we get

\left \{ {p+d+118+d=273} \atop {6p+4d+2d+236=836}} \right.

\left \{ {{p+2d=115} \atop {6p+6d=600}} \right.

Now we get from the upper one

p=115-2d

solving the another equation gives us

6*115-12d+6d=600,

hence d=15

and by replacing

p=115-2*15=85.

85 premium tickets were sold

lesya692 [45]3 years ago
3 0

Answer:

45 premium tickets were sold

Step-by-step explanation:

p = premium

d = deluxe

r = regular

p+d+r = 273

6p+4d + 2r = 836

118+d = r

Replace r with 118+d

p+d+118+d = 273

p +2d = 273-118

p+2d = 155

6p+4d + 2(118+d) = 836

6p+4d + 236+2d = 836

6p +6d = 836-236

6p + 6d = 600

Divide by 6

p+d = 100

d = 100-p

Replace d in p +2d= 155

p +2(100-p) = 155

p+200-2p = 155

-p = 155-200

-p =-45

p =45

45 premium tickets were sold

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A percentage is a number out of 100.  So, 17.5% is really 17.5/100, or 0.175.

To find 17.5% of 1500, you multiply 1500 by 0.175

1500 x 0.175 = 262.5.

The commission is $262.50.
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Find the perimeter of a square equal in area to two squares whose sides are 12 m and 5 m respectively
KatRina [158]
Formula for perimeter a+b+a+b

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3 years ago
The American Psychological Associated conducted a survey in 2003 of psychologists to estimate the average incomes for psychologi
Alex Ar [27]

This question is incomplete, the complete question;

The American Psychological Associated conducted a survey in 2003 of psychologists to estimate the average incomes for psychologists based on their academic degree and years of experience. 36 psychologists responded who had a just received their masters degree in 2003. The average was $48,800 with a standard deviation of $16,800.

The 95% Confidence Interval is ($, $ )

Answer:

at 95%, confidence interval is ( $43312, $54288 )

Step-by-step explanation:

Given the data in the question;

sample size n = 36

mean x' = 48,800

standard deviation σ = 16,800

at 95% confidence level;

∝ = 1 - 95% = 1 - 0.95 = 0.05

∝/2 = 0.05 / 2 = 0.025

z critical value, Z_{\alpha/2 = 1.96

now, confidence interval for population mean is given by the formula

x' ±  Z_{\alpha/2( σ/√n )

so we substitute

⇒ 48800 ±  1.96( 16800/√36 )

⇒ 48800 ±  1.96( 2800 )

⇒ 48800 ±  5488

⇒ 48800 -  5488, 48800 +  5488

⇒ ( 43312, 54288 )

Therefore, at 95%, confidence interval is ( $43312, $54288 )

6 0
2 years ago
A recent survey indicated that the average amount spent for breakfast by business managers was $9.33 with a standard deviation o
olasank [31]

Answer:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

Step-by-step explanation:

Information given

\bar X=9.41 represent the sample mean

s=0.24 represent the sample standard deviation

n=81 sample size  

\mu_o =9.33 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is higher than 9.33, the system of hypothesis would be:  

Null hypothesis:\mu \leq 9.33  

Alternative hypothesis:\mu > 9.33  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

4 0
3 years ago
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