If m ∥ k and m ∥ ℓ, then _____

Mathematics

1 answer:

saul85 [17]2 years ago

4 0

Answer:

$\large\huge\boxed{\text{If}\ m\ ||\ k\ \text{and}\ m\ ||\ l,\ \text{then}\ k\ ||\ l}$

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Kiara's birthday celebration started at 2:00 pm. She spent 46 minutes at the trampoline park, Eating cake and opening presents t

Phantasy [73]

Answer:

Im not 100% sure but I got 1.66667 so therefore I would say about 4:00 maybe? Didn't really understand lmk.

Step-by-step explanation: I added 46, 29, and 25 together which gave me 100 and i put that into minutes which gave me 1.66667 which rounds to 1.70 and so if 60 seconds is 1 hour than 2:00 plus 2 hours is 4. That's what I got out of it.

5 0

2 years ago

A certain change in a process for manufacture of component parts is being considered. Samples are taken using both the existing

Katyanochek1 [597]

Answer:

a) The confidence interval is $0.00\leq\pi_1-\pi_2\leq0.02$ .

Step-by-step explanation:

We have to calculate a confidence interval (CI) of a difference of proportions.

For the existing procedure, the proportion is:

$p_1=75/1500=0.05$

For the new procedure, the proportion is:

$p_2=80/2000=0.04$

To calculate the CI, we need to estimate the standard deviation

$s=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2} }\\\\s=\sqrt{\frac{0.05(1-0.05)}{1500} +\frac{0.04(1-0.04)}{2000} }=\sqrt{ 0.000032 + 0.000019 }=\sqrt{ 0.000051 }\\\\s=0.007$

For a 90% CI, the z-value is 1.64.

Then, the CI is:

$(p_1-p_2)-z*\sigma \leq\pi_1-\pi_2\leq(p_1-p_2)+z*\sigma \\\\(0.05-0.04)-1.64*0.007\leq\pi_1-\pi_2\leq(0.05-0.04)+1.64*0.007\\\\0.01-0.01\leq\pi_1-\pi_2\leq0.01+0.01\\\\0.00\leq\pi_1-\pi_2\leq0.02$