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Firlakuza [10]
3 years ago
10

PLEASE HELP ME I NEED HELP

Mathematics
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

13 is c

Step-by-step explanation:

to find the  length of a diagonal, pretend as if you are finding the hypotenuse of a triangle.

5^2 + 12^2 = c^2

25 + 144 = 169

the square root of 169 is 13

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What can u show pic

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6. What is the greatest common factor of 12wr2 + 6wr – 6w ?
balandron [24]

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a

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Find f-1(x), if f(x) = 3x – 6. Justify your
Sergio039 [100]
Answer: 1/3 (x + 6)

Explanation: given f(x) = 3x - 6.
Let y = f(x) and so y = 3x - 6.
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x = 3y - 6
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4 0
2 years ago
I really need the answer
notka56 [123]

Answer:

2 \frac{4}{16} = 2 \frac{2}{8} = 2 \frac{1}{4} lb.

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
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:

<em>The question is incomplete:</em>

<em>"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."</em>

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

8 0
3 years ago
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