a) These samples are dependent, as measurements are taken from the same individuals at different times using different methods. b) Upon examining the QQ plot, it shows no considerable deviations suggesting a normal distribution assumption for the differences. c) For the hypotheses indicated, a paired t-test is suitable due to the repeated measurements. Evaluating the p-value reveals it exceeds the significance threshold, leading to the conclusion that we FAIL to reject the null hypothesis, thus indicating that the mean difference is not significantly different from 0.
a) P(identified as explosive) equals P(actual explosive & identified as explosive) + P(not explosive & identified as explosive) = (10/(4*10^6))*0.95+(1-10/(4*10^6))*0.005 = 0.005002363. Thus, the probability that it actually contains explosives given that it's identified as containing explosives is (10/(4*10^6))*0.95/0.005002363 = 0.000475. b) Let the probability of correctly identifying a bag without explosives be a. Therefore, a = 0.99999763, approximately 99.999763%. c) No, even if this becomes 1, the true proportion of explosives will always be below half of the total detected.
The capacity is approximately 3.5 fluid ounces. In order to determine this, we need to compute the volume of a cone-shaped cup. The formula for the volume of a cone is: 1/3 * π * r^2 * h, where r equals the diameter divided by 2, which gives 1.35 inches, and h equals 3.3 inches. After substituting these values, we find the volume to be V = 1/3 * 3.14 * 1.35^2 * 3.3 = 6.3 cubic inches. To convert cubic inches to fluid ounces, we use the relationship that 1 fluid ounce is equal to 1.8 cubic inches. Therefore, x fluid ounces equal to 6.3 cubic inches leads to x = 6.3 / 1.8, which results in 3.5 fluid ounces.
Answer:
one hour.
Step-by-step explanation:
by subtracting 24 from 36, you arrive at 12, which when multiplied by 5 gives you 60. 60 minutes equals one hour.
hope this is useful!!
apologies if incorrect..
