What do we know about these angles? Immediately, you might notice that (4y-8)° and (16x-4)° share a line. The same is true of (16x-4)° and (14x+4)°. Any straight line forms what's called a <em>straight angle</em>, which measures 180°, so we know that, since they add up to form a straight angle, (14x+4)° and (16x-4)° must add up to 180°. We can use that fact to set up an equation to solve for x:
(14x+4)+(16x-4)=180
After you solve for x, you should look to solve for y. How can we figure out what y is? If you're familiar with the vertical angle theorem, you'll know that all vertical angles (angles that are directly across from each other diagonally) are equal. So we know that 14x+4=4y-8. You can use the value of x you solved for before to solve this one fairly easily, and then you'll have both values.
If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days
Answer:
a_23 = 5
Step-by-step explanation:
The row is the first number so in a_23 The row is a 2
The column is the second number so the column is 3
The second row is 3 9 5 1
The third column number would be 5
The answer is a
Answer: its 4
Step-by-step explanation:
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Answer:
- <u><em>Option D. 0.50</em></u>
Explanation:
1. Two-way table:
Calories per Day:
1000 to 1500 1500 to 2000 2000 to 2500 Total
Weight
120 lb. 90 80 10 180
145 lb. 35 143 25 203
165 lb. 15 27 75 117
Total 140 250 110 500
2. Total number of persons
Look at the intersection of the totals for the columns and the rows: 500
3. Number of persons that consume 1,500 to 2,000 calories in a day
Look at the total for the column 1,500 to 2,000 calories per day: 250
4. Probability that a person consumes 1,500 to 2,000 calories in a day, P
- P = number of persons that consume 1,500 to 2,000 calories / total number of persons
- P = 250 / 500 = 0.50 ↔ option D.
