When constructing the histogram, the intervals are selected in such a way that the width of each interval is the same and the gap between two intervals stays uniform throughout the histogram. This makes the data easy to visualize, easy to interpret and easy to understand.
From the given options, option B lists the intervals in an absurd order. First interval is from 0 to 3. There is no gap between first and second interval and the second interval is larger than the first. So these intervals cannot be the set of intervals of a histogram.
So, the answer to this question is option B
Use the lists of multiples to answer the problem. Multiples of 8: 8, 16, 24, 32, 40, . . . Multiples of 12: 12, 24, 36, 48, 60,
patriot [66]
Answer:
24 minutes
Step-by-step explanation:
Look at the lists of multiples and find the first instance of the same number appearing in both lists.
Multiples of 8: 8, 16, 24, 32, 40, ...
Multiples of 12: 12, 24, 36, 48, 60, ...
The first number that appears in both lists is 24, so the answer is 24.
Answer: 24 minutes
Answer: 3/4 i think
Step-by-step explanation:
Answer:
cos3x+tan3x=0
⟹cos3x=−tan3x
⟹cos3x=−sin3xcos3x
⟹cos23x=−sin3x
⟹1−sin23x=−sin3x
⟹sin23x−sin3x−1=0
This is a quadratic equation in sin3x.
sin3x=−(−1)±(−1)2−4×1×(−1)−−−−−−−−−−−−−−−−−√2×1
sin3x=1±5–√2
If x takes real values, the upper sign must be rejected.
sin3x=1−5–√2
⟹3x=nπ+(−1)nsin−11−5–√2
⟹x=13[nπ+(−1)nsin−11−5–√2]
Step-by-step explanation:
Hope this kind of helps
We can't help unless we know what to "rewrite"
