Will mark brainliest!
show work.

h(x) has the smallest minimum.

Step-by-step explanation:

f(x) = -4sin(x-0.5) + 11

y = sin(x – 0.5) has minima at y =  -1.

y = -4sin(x – 0.5) has minima at y = -4.

y = sin(x – 0.5) + 11 has minima at y = 7.

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According to your graph,

g(x) has a minimum at y = 6.

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According to your table,

h(x) has a minimum at y = 5.

Thus, h(x) has the smallest minimum.

The graph below shows the minima of f(x) at y = 7 and the minimum of h(x) at y = 5.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]