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Ber [7]
3 years ago
11

David went to a lot of concerts over the past year with different groups of friends. He went to a concert that cost $32 per tick

et with one friend, a concert that cost $18 per ticket with two friends, and a concert that cost $54 by himself. How much money was spent by David and his friends on tickets?
Mathematics
1 answer:
atroni [7]3 years ago
3 0
David spent $172.

The first concert he went with 1 friend including himself so
32+32=64
The second concert he went with 2 friends including himself so
64+18+18+18=118
The third concert he went by himself so
118+54=172
David spent $172 dollars with him and his friends on tickets.
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Answer: B.) Find the difference between the upper and lower quartiles.

Step-by-step explanation: Edgen 2020

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5 0
4 years ago
There are 90 girls and 60 boys in the 6th grade at a middle school. Of these students, 9 girls and 3 boys write left-headed. Wha
natta225 [31]

Answer:

8%

Step-by-step explanation:

9+3=12 (total number of students that write left handed)

90+60=150 (total number of 6th graders)

12/150 x 100% = 8%

6 0
2 years ago
Each year, all final year students take a mathematics exam. It is hypothesised that the population mean score for this test is 1
Yuri [45]

Answer:

90% confidence interval for the population mean test score is [95.40 , 106.59]

Step-by-step explanation:

We are given that the population mean score for mathematics test is 115. It is known that the population standard deviation of test scores is 17.

Also, a random sample of 25 students take the exam. The mean score for this group is 101.

The, pivotal quantity for 90% confidence interval for the population mean test score is given by;

        P.Q. = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean = 101

              \sigma = population standard deviation

              n = sample size = 25

So, 90% confidence interval for the population mean test score, \mu is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.6449) = 0.90

P(-1.6449 * \frac{\sigma}{\sqrt{n} } < {Xbar-\mu} < 1.6449 * \frac{\sigma}{\sqrt{n} } ) = 0.90

P(X bar - 1.6449 * \frac{\sigma}{\sqrt{n} } < \mu < X bar + 1.6449 * \frac{\sigma}{\sqrt{n} } ) = 0.90

90% confidence interval for \mu = [ X bar - 1.6449 * \frac{\sigma}{\sqrt{n} } , X bar + 1.6449 * \frac{\sigma}{\sqrt{n} } ]

                                                  = [ 101 - 1.6449 * \frac{17}{\sqrt{25} } , 10 + 1.6449 * \frac{17}{\sqrt{25} } ]

                                                  = [95.40 , 106.59]

Therefore, 90% confidence interval for the population mean test score is [95.40 , 106.59] .

7 0
4 years ago
Sz – [32 – (10z + 2)]​
Dmitry_Shevchenko [17]
Sz-[32-(10z+2)]
=Sz-(32-10z-2)
=Sz-(30-10z)
=Sz-30+10z

Hope my answer helped u :)
5 0
3 years ago
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