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2 years ago

12

Which number is a common factor of 50 and 125? A) 15 B) 25 C) 45 D) 50 F) Jk, I'm just joking with you guys, anyone who answers

first can be brainliest

1 answer:

oksian1 [2.3K]2 years ago

6 0

Answer:

The answer is *drum roll* ... 25!

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How do you turn a fahrenheit to kelvins??

densk [106]

<span>1.Subtract 32 from the Fahrenheit temperature.
2.Multiply this number by 5.
3.Divide this number by 9.
<span>4.Add 273.15 to this number.</span></span>

6 0

2 years ago

The average monthly rainfall for 6 months was 28.5 mm. If it had rained 1mm more each month what would the average have been? By

Fynjy0 [20]

Answer:

29.5 mm

Step-by-step explanation:

We are told that the average monthly rainfall for 6 months was 28.5 mm.

Thus, total for the 6 months = 28.5 × 6 = 171 mm

Now, we are told that it rained 1 mm extra each month.

So extra for the six months = 1 × 6 = 6mm

New total for 6 months = 171 + 6 = 177 mm

So, new average for 6 months = 177/6 = 29.5 mm

4 0

2 years ago

Yep here we go again:

Alchen [17]

Answer:

126

Step-by-step explanation:

360 x 35%= 126

360 x 0.35 = 126

3 0

2 years ago

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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

I know the answer is A....but why?

katrin2010 [14]

Answer:a

Step-by-step explanation:

throughout the table it is presented that the answer, A, is correct becuase within the percentages is is obvious that a is correct.

6 0

3 years ago

Read 2 more answers