If m <NKJ =5x+19 and m<4=x, what is m<4?
1 answer:
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<u><em>Answer:</em></u>
m∠4 = 23°
<u><em>Explanation:</em></u>
<u><em>1- getting the value of x:</em></u>
<u>We know that KM bisects ∠NKL, this means that:</u>
∠3 = ∠4 = x
Now, we know that ∠NKJ, ∠3 and ∠4 form <u>a straight angle</u>.
This means that their summation is 180°
<u>We can now get the value of x as follows:</u>
180 = ∠NKJ + ∠3 + ∠4
180 = 5x + 19 + x + x
180 = 7x + 19
180 - 19 = 7x
161 = 7x
x = 23
<u><em>2- getting ∠4:</em></u>
From part 1, we got that x = 23
We are given that:
∠4 = x
This means that:
∠4 = 23°
Hope this helps :)
<u><em>Answer:</em></u>
m∠4 = 23°
<u><em>Explanation:</em></u>
<u><em>1- getting the value of x:</em></u>
<u>We know that KM bisects ∠NKL, this means that:</u>
∠3 = ∠4 = x
Now, we know that ∠NKJ, ∠3 and ∠4 form <u>a straight angle</u>.
This means that their summation is 180°
<u>We can now get the value of x as follows:</u>
180 = ∠NKJ + ∠3 + ∠4
180 = 5x + 19 + x + x
180 = 7x + 19
180 - 19 = 7x
161 = 7x
x = 23
<u><em>2- getting ∠4:</em></u>
From part 1, we got that x = 23
We are given that:
∠4 = x
This means that:
∠4 = 23°
Hope this helps :)
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Answer:
Mean track length for this rock specimen is between 10.463 and 13.537
Step-by-step explanation:
99% confidence interval for the mean track length for rock specimen can be calculated using the formula:
M± where
- M is the average track length (12 μm) in the report
- t is the two tailed t-score in 99% confidence interval (2.977)
- s is the standard deviation of track lengths in the report (2 μm)
- N is the total number of tracks (15)
putting these numbers in the formula, we get confidence interval in 99% confidence as:
12± =12±1.537
Therefore, mean track length for this rock specimen is between 10.463 and 13.537