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Oxana [17]
3 years ago
15

A House needs to some maintenance one sink is dripping water every 10 minutes and the other sink is dripping every 12 minutes if

both sink dripped at the same time how many minutes until they both drip at the same time again
Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

60 minutes

Step-by-step explanation:

We solve this question using Lowest Common Factor method

Find and list multiples of each number of minutes( 12 and 10 minutes) until the first common multiple is found. This is the lowest common multiple.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80

Multiples of 12:

12, 24, 36, 48, 60, 72, 84

Therefore,

LCM(10, 12) = 60

The number if minutes until they both drip again is 60 minutes

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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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3 years ago
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