Answer:
Step-by-step explanation:
tax = 1/20 of price
1/20 of $3 = 1/20 × 3 = 3/20 = $0.15
1/20 of $800 = 1/20 × 800 = $40
Answer:
B. 0 ≤ x ≤ 6
Step-by-step explanation:
The horizontal grid lines are 10 °C apart, so 45 °C corresponds to a vertical position halfway between the 4th grid line and the one marked 50.
The graph intersects that position at the 2nd vertical grid line. The vertical grid lines are 3 minutes apart, so the 2nd grid line represents 6 minutes.
If Peter conducted the experiment from its start until the temperature reached 45 °C, he conducted it over a period of 0 to 6 minutes, inclusive:
0 ≤ x ≤ 6
Answer:
<h2>√10</h2>
Step-by-step explanation:
![\text {modulus of z } = =\sqrt{\left( -3\right)^{2} +\left( 1\right)^{2} } =\sqrt{9+1} =\sqrt{10}](https://tex.z-dn.net/?f=%5Ctext%20%7Bmodulus%20of%20z%20%7D%20%3D%20%3D%5Csqrt%7B%5Cleft%28%20-3%5Cright%29%5E%7B2%7D%20%20%2B%5Cleft%28%201%5Cright%29%5E%7B2%7D%20%20%7D%20%3D%5Csqrt%7B9%2B1%7D%20%3D%5Csqrt%7B10%7D)
Answer:
945 muffins
Step-by-step explanation:
135 * 7 = 945
Answer:
a) 2.5% b) 84% c) 95% d) D. The more unusual day is if the stock closed below $185 because it has the largest absolute z-score.
Step-by-step explanation:
For a) b) and c) we will use the empirical rule, so, we can observe the image shown below
a) 211.23 is exactly two standard deviation above the mean, so, the probability that on a randomly selected day in this period the stock price closed above 211.23 is 2.35% + 0.15% = 2.5%
b) 204.11 represents exactly one standard deviation above the mean, so, the probability of being below 204.11 is 50% + 34% = 84%
c) The probability of getting a value between 182.75 and 211.23 is 95%, this because 182.75 is exactly two standard deviations below the mean and 211.23 is exactly two standard deviations above the mean.
d) The z-score related to 208 is
= (208-196.99)/7.12 = 1.5 and the z-score related to 185 is
= (185-196.99)/7.12 = -1.7, therefore, the more unusual day is if the stock closed below $185 because it has the largest absolute z-score.