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Vedmedyk [2.9K]
3 years ago
11

PLZZZ HELP ME ITS DUE TODAY AT 11:59, HELPPPP

Mathematics
2 answers:
Andrew [12]3 years ago
8 0

Answer:

1. 135

2. 21

3.-10

Step-by-step explanation:

kirill [66]3 years ago
4 0

Answer:

1) . Divide that number by 12 months

1620/12 = 135

2) 3 x 7 = 21

21 pounds lost

3) 0 - 4 - 4 - 4 + 2 = -10 points

hope this helps :)

Step-by-step explanation:

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15 POINTS!!!
MaRussiya [10]

Answer:

1. The expected pay-out on each policy is 250 * 1/90 + 12000 * 1/100 + 17000 * 1/400 = $165.  So that's what the premium would have to be in order to get a profit of 0.

 

2. The profit per policy is the premium the company receives minus the expected payout = 350 - 165 = $185.

 

3. The expected profit on 375 policies would be 375 * 185 = $69375

Step-by-step explanation:

7 0
3 years ago
54x^3 +250y^3<br> in the sum of cubes<br> please helpppp
Slav-nsk [51]

Answer:

What is exactly the question?

4 0
3 years ago
What is the factor to change pounds to kilograms
aev [14]

Answer:

1 pounds is equal to 4535 kilograms.

8 0
3 years ago
What is the product?
RSB [31]

Answer:

15a^5b^15

Step-by-step explanation:

you multiply 5 and 3 to get 15, and then when you are multiplying exponents of variables, you actually add them, so a^2 times a^3 is actually a^5, and b^7 times b^8 is b^15. Because these are all multiplied together, you get the answer of 15a^5b^15.

4 0
3 years ago
Read 2 more answers
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
3 years ago
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