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blsea [12.9K]
3 years ago
7

Jarek buys jerseys for his team online. He pays a constant shipping price plus a special rate for each jersey. During the spring

season, Jarek paid $151 for 24 jerseys. In the summer season, he paid $79 for 12 jerseys. What is the special rate Jarek pays for each jersey and how much does he pay for shipping?
Thanks to anyone who can help!
Mathematics
1 answer:
Anika [276]3 years ago
3 0
The rate is $6 and the shipping is $7
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Please help! Could you answer all!?
Brums [2.3K]

Answer:

1.) 48

2.) 65

3.) 36

Step-by-step explanation:

1.)     If the equation is 6(x-4) and x = 12, then all we have to do is plug in the value of x. When we plug in, all we do is substitute 12 for x because they mentioned in the question that x = 12. So, we end up getting 6(12 - 4). After solving this, we get 48.

2.)     This problem is a lot like the last problem. All we need to do is substitute /plug in the values of x and y into the equation, to get 4(4^2) - 35/7 - (8 + 14). After solving, we get 65.

3.) .     This problem, once again, is also a lot like the last problems. We need to substitute the value of x into the equation 8x+12. Since we know from the problem that x is 3, all we have to do is 8 * 3 + 12.

3 0
3 years ago
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Linear equations: w-3=15
motikmotik

Answer:

w = 18

Step-by-step explanation:

w-3=15

Add 3 to each side

w-3+3=15+3

w = 18

6 0
3 years ago
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A​ 12-sided die is rolled. The set of equally likely outcomes is​ {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling
UNO [17]
1/6 is the probability of rolling a dice and getting less than 3.
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3 years ago
1+4(6p-9) simplified
snow_lady [41]

Answer

24p+ (-31)

Step-by-step explanation:

  1. P.EM.D.A.S
  2. 4*6p=24p
  3. 4*-9=-36
  4. now it looks like: 1+4+24p+-36
  5. now for addition 1+4+24p+-36= 24p+ (-31)
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Berto invested in a precious mineral. The value of the mineral tend to increase by about 11% per year. He invest $40,000 in 2015
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the difference from 40000 to the future value in 2020, namely 5 years later is simply the yield or interest amount.


\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$40000\\ r=rate\to 11\%\to \frac{11}{100}\dotfill &0.11\\ t=years\dotfill &5 \end{cases} \\\\\\ I=(40000)(0.11)(5)\implies I=22000

5 0
3 years ago
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