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3 years ago

Find two numbers for which the sum is 101 and the difference is 47

Mathematics

2 answers:

xenn [34]3 years ago

8 0

74 and 27

74+27=101
74–27=47

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ANEK [815]3 years ago

5 0

Answer:

74 and 27

Step-by-step explanation:

let x and y be the numbers

x + y =101........eqn 1

x - y = 47.......eqn 2

solve simultaneously

from equation 2, make x the subject

x= 47 + y........eqn 3

put eqn 3 into eqn 1

(47+y) + y = 101

47 + 2y = 101

2y= 101 - 47

2y=54

y= 54/2

y= 27

put y=27 into eqn 3

x = 47 + 27

x = 74

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2 years ago

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

The question is incomplete:

"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$