A person with type A blood can donate red blood cells to people with type A or type AB blood. About 31% of the US population has type A blood. You are asked to design a simulation to determine the probability that at least 4 of 10 people will have type A blood. Part A: Conduct five trials using the following table of random digits:

96299 07196 98642 20639 23185 56282 69929 14125 38872 94168
71622 35940 81807 59225 18192 08710 80777 84395 69563 86280
03272 41230 81739 74797 70406 18564 69273 72532 78340 36699
46376 58596 14365 63685 56555 42974 72944 96463 63530 24152
47352 04266 74017 79319 70170 96572 08523 56025 89077 57678

Part B: Based on your simulation, what is the probability that at LEAST 4 of 10 people will have type A blood?

Answerpeople,vehicles,roadways.;

Answer:     the interval on which the curve is concave up is  (-∞, -1/2)======    we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x    let the function f denote be the anti-derivative of the inside expression, i.e. a function such that .then use the fundamental theorem of calculus:     differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function    find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x    where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation    1 + x + x² = 0results in a discriminant of      meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the  curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is  (-∞, -1/2)