Bongiwe invests R12 000 in a savings account at 6,5% per annum compound interest. Calculate how much there will be in the saving
1 answer:
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Answer:
0.2109 or 21.09%
Step-by-step explanation:
In order to maintain the same price after two days, the stock must go up (U) on two days and go down (D) on two days, the sample space for this event is:
S={UUDD, UDUD, UDDU, DDUU, DUDU, DUUD}
There are 6 equally likely possible outcomes. The probability that the price of the stock will be the same as it is today is:
The probability is 0.2109 or 21.09%.
Answer:
There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.
Step-by-step explanation:
We notice that the expression on the left of the equation is a quadratic with leading term , which means that its graph is that of a parabola with branches going up.
Therefore, there can be three different situations:
1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.
2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.
3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.
So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.
We recall that the x-position of the vertex for a quadratic function of the form is given by the expression:
Since in our case and
, we get that the x-position of the vertex is:
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:
This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.
We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:
See the graph produced in the attached image.
Answer:
≈50.6
Step-by-step explanation:
Not sure what precision level this problem is looking for, but for right-skewed distributions, we know that the mean is going to be pulled right and therefore the mean should be larger than the median. To a high confidence level, the mean should fall between 50 and 59, or in the third column.
If a single estimation is wanted, assume the values inside each column are uniformly distributed: