Ask question

Ask question

All categories

2 years ago

13

In a certain mid-size city, 90% of the high- school graduates are from public schools and the other 10% are from private schools

. Of the private school graduates, 80% go on to complete college but only 50% of the public school graduates complete college. What is the probability that a high school graduate from this city will complete college?
A. 0.53
B. 0.58
C. 1.3
D. 0.9
E. 0.47

1 answer:

trapecia [35]2 years ago

8 0

The answer should be .53

You might be interested in

What is the remainder in the synthetic division problem below?

Zepler [3.9K]

Answer:

<h2>D.8</h2>

Step-by-step explanation:

The synthetic division is

| 4 6 -2

1 | 4 10

4 10 8

As you can observe, the remainder in the given synthetic division is 8, because that's the ultimate result.

Therefore, the right answer is D.8

5 0

2 years ago

The widths (in meters) of a kidney-shaped swimming pool were measured at 3-meter intervals as indicated in the figure. Use the M

forsale [732]

We can used the Simpson's Rule says to approximate the area under a given curve using the following formula:

<span>(Δx/3)[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + ... + 2f(xn-2) + 4f(xn-1) + f(xn)] </span>

<span>The pool is divided into 8 subintervals. We integrate the given function from 0 to 24, while the graph provides values of f(x) at 7 different points. The first value given, 6.2, is NOT f(0). It is f(3). Using Simpson's Rule, and dividing the lake of 24 meters into 8 subintervals, we write the equation: </span>

<span>area = (3/3)[f(0) + 4f(3) + 2f(6) + 4f(9) +2f(12) + 4f(15) + 2f(18) + 4f(21) + f(24)] </span>

<span>Pool area = 0 + 4(6.2) + 2(7.2) +4(6.8) + 2(5.6) + 4(5.0) +2(4.8) +4(4.8) + 0 = 126.4 m^2 </span>

<span>Rounding to the nearest square meter, the area of the lake is approximately 126 m^2 </span>

6 0

2 years ago

Read 2 more answers

Find the point (x,y) of x2+14xy+49y2=100 that is closest to the origin and lies in the first quadrant.

Gala2k [10]

Notice that

$x^2+14xy+49y^2=(x+7y)^2$

so the constraint is a set of two lines,

$(x+7y)^2=100\implies\begin{cases}x+7y=10\\x+7y=10\end{cases}$

and only the first line passes through the first quadrant.

The distance between any point $(x,y)$ in the plane is $\sqrt{x^2+y^2}$ , but we know that $\sqrt{f(x,y)}$ and $f(x,y)$ share the same critical points, so we need only worry about minimizing $x^2+y^2$ . The Lagrangian for this problem is then

$L(x,y,\lambda)=x^2+y^2+\lambda(x+7y-10)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0$

$L_y=2y+7\lambda=0$

$L_\lambda=x+7y-10=0$

We have

$L_y-7L_x=2y-14x=0\implies y=7x$

which tells us that

$x+7y-10=0\iff x+49x=10\implies x=\dfrac15\implies y=\dfrac75$

so that $\left(\dfrac15,\dfrac75\right)$ is a critical point. The Hessian for the target function $x^2+y^2$ is

$H(x,y)=\begin{bmatrix}2&0\\0&2\end{bmatrix}$

which is positive definite for all $x,y$ , so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is $\sqrt{\left(\dfrac15\right)^2+\left(\dfrac75\right)^2}=2$ .

3 0

2 years ago

A karate center has 120 students. The director wants to set a goal to motivate her instructors to increase student enrollment. U

iVinArrow [24]

A) Plan A requires for a percentage increase of a number of students. This means that year after year the number of new students will increase. Plan B requires for a constant number of new students each year. This means that year after year the percentage increase would get smaller.

B) To solve this problem we will use formula for a growth of population:
$final = initial * (1+percentage)^{t}$
Where:
final = final number of students
initial = initial number of students
percentage = requested percentage increase
t = number of years

We can insert numbers and solve for t:
$240= 120* (1+0.12)^{t} \\ 2=1.12^{t} \\ log2=log(1.12^{t} ) \\ log2=t*log1.12 \\ t= \frac{log2}{log1.12} \\ t=6.12years$

For Plan B we can use simple formula
increase = 120
increase per year = 20
number of years = increase / (increase per year) = 120 / 20 = 6 years

Plan B is better to double the <span>enrollment.

C)We use same steps as in B) to solve this.
</span>

$360= 120* (1+0.12)^{t} \\ 3=1.12^{t} \\ log3=log(1.12^{t} ) \\ log3=t*log1.12 \\ t= \frac{log3}{log1.12} \\ t=9.69years$

For Plan B we can use simple formula
increase = 240
increase per year = 20
number of years = increase / (increase per year) = 240 / 20 = 12 years

Plan A is better to triple the enrollment.

3 0

2 years ago

Which ordered pair shows an equivalent ratio of hours to miles?

ANEK [815]

8,80 ywahhh ok yw trying to get 20 scharcters

3 0

1 year ago

Read 2 more answers