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BaLLatris [955]
3 years ago
14

The sum of three numbers is 18. If every number is a prime number, what are the three numbers?

Mathematics
2 answers:
Andru [333]3 years ago
5 0
2,5,11 because every number is a prime and adding thoose up equal 18
sleet_krkn [62]3 years ago
4 0
The three numbers are 2, 5, and 11.

The sum of 3 numbers is 18.        2 + 5 + 11 = 18

Since the problem states that every number is a prime number, then the number must be a natural number greater than 1 that has no positive divisors other than 1 and itself.

2 / 1 = 2  OR  2 / 2 = 1  These numbers are prime numbers because their divisors
5 / 1 = 5  OR  5 / 5 = 1            are 1 and itself.
11/1 = 11 OR 11/11 = 1
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Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz
GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender   Mean          Std. dev.     n

Female    2577.75      1916.29     67

Male        3809.42     2379.47     33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

M_d=M_1-M_2=2577.75-3809.42=-1231.67

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=1.96 \cdot 475.795=932.54

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

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