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1 year ago

What is the least common multiple of 36, 35, and 11?

Mathematics

1 answer:

xz_007 [3.2K]1 year ago

6 0

Answer:

13,860

Step-by-step explanation:

List all prime factors for each number.

Prime Factorization of 11 shows:

11 is prime.

Prime Factorization of 35 is:

5 x 7

Prime Factorization of 36 is:

2 x 2 x 3 x 3

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new superset list is

2, 2, 3, 3, 5, 7, 11

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 3 x 3 x 5 x 7 x 11 = 13860

LCM(11, 35, 36) = 13,860

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The interval is $-0.0037 < p_1-p_2$

Step-by-step explanation:

From the question we are told that

The first sample size is $n _1 = 1068000$

The first sample proportion is $\r p_1 = 0.089$

The second sample size is $n_2 = 1476000$

The second sample proportion is $\r p_2 = 0.092$

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$\alpha = (100 - 95 )\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

The value is

$Z_{\frac{\alpha }{2} } =z_c= 1.96$

Generally the 95% confidence interval is mathematically represented as

$(\r p_1 - \r p_2 ) -z_c \sqrt{ \frac{\r p_1 \r q_1 }{n_1} + \frac{\r p_2 \r q_2 }{n_2}} < (p_1 - p_2 ) < (\r p_1 - \r p_2 ) +z_c \sqrt{ \frac{\r p_1 \r q_1 }{n_1} + \frac{\r p_2 \r q_2 }{n_2}}$

Here $\r q_1$ is mathematically evaluated as $\r q_1 = (1 - \r p_1)= 1-0.089 =0.911$

and $\r q_2$ is mathematically evaluated as $\r q_2 = (1 - \r p_2) = 1- 0.092 = 0.908$

$(0.089 - 0.092 ) -1.96 \sqrt{ \frac{0.089* 0.911 }{1068000} + \frac{0.092* 0.908 }{1476000}} < (p_1 - p_2 ) < (0.089 - 0.092 ) +1.96 \sqrt{ \frac{0.089* 0.911 }{1068000} + \frac{0.092* 0.908 }{1476000}}$

$-0.0037 < p_1-p_2$

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