- (A) The following data was received from the Student Affairs Unit of the University

Mathematics

1 answer:

ladessa [460]3 years ago

5 0

Answer: i) 30 students for chemistry

ii) 21 students for only one subject

iii) 61 students

Step-by-step explanation: from the question we got 30 students registered for chemistry. number students only for biology = (27-18) = 9

number of students only for chemistry = (30-18) = 12

number of student registered for only one subject = (12+9)= 21

number of students registered for neither two subjects = {100-(9+12+18)} = 61

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125. Albert surveyed a class of 25 students on sports. 5 kids love baseball. 7 kids love basketball. 10 kids

MAXImum [283]

Answer:

3 students

Step-by-step explanation:

since the total number of students is 25,when you add those that like baseball, basketball and football the total number must be 25 but in this case it's 22 meaning 2 student liked neither.

7+5+10+x=25

x=25-22

I hope this helps

4 0

3 years ago

The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func

alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

$h(t)=-16t^2 + 96t$

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

$To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)$

$Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution$