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2 years ago

Suppose 360 nursing students took a certification test...

Mathematics

2 answers:

Leya [2.2K]2 years ago

4 0

If the third quartile (Q3) is 130, this means that around 3/4 of the scores were below or equal to 130. Out of the 360 students who took the exam:
(3/4)(360 students) = 270 students who got scores of 130 or less
This is the last of the given choices.

Semenov [28]2 years ago

3 0

Out of 360 nursing students who took the test, 270 students got a score of 130 or lesser

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For the sequence below, explain the pattern in words. 14. 11. 8. 5....

AVprozaik [17]

Answer:

you subtract the number by 3

Step-by-step explanation:

the pattern is x-3 so the pattern is subtracting the number by 3 every time

6 0

2 years ago

If f(x)=x^2-7 and h(x)=7x-5, then f(h(2))=

zaharov [31]

We are given 3 equations. Using these equations we can find f(h(2)).

h(x)=7x-5
h(2)=7(2)-5
h(2)=14-5
h(2)=9

f(x)=x^2-7
f(h(2))=(7x-5)^2-7
we know the answer for h(2) by now, so we can substitute for that.
f(9)=9^2-7
f(9)=81-7
f(9)=74

so f(h(2))=74

5 0

3 years ago

| 9. The distance between Town A and Town B was 108 km. A car and a van left Town A at 12 00 for Town B. On reaching Town B, the

Westkost [7]

Answer:

a) Time until both vehicles meet is 1.5 hours after starting at noon. That makes it 1:30PM.

b) Average speed of car is 84 km/h

Step-by-step explanation:

A -----------------------z------------B

Left Speed(km/h) Time

Car: 12PM X

Van: 12PM 60

Car/Van

DistanceCar AB + z

DistanceVan Az

Ratio: (AB+z)/Az = 7/5

Time until both meet = T (in hours)

Distance Car: xT

Distance Van: 60T

====

xT = AB + z

60T = Az

---

(xT/60T)= (7/5)

x = 60(7/5)

x = 84 km/h

=====

Time for car to reach B is: time (hr) = 108 km/(84 km/h)

time = 1.286 hours

Distance for at 1.289 hours is: distance (km) = (60 km/h)*(1.286 h)

distance = 77.14 km

At 1.286 hours, the car reverses direction. The van is (108 km - 77.14 km) or 30.86 km away.

Add the distances travelled by both vehicles after the car reverses direction at 1.286 hours. The sum will be 30.86 km when they meet, at time of T.

Car Distance + Van Distance = 30.86 km

T(84 km/h) + t(60 km)

They meet when they are 0 km apart, which can be modeled with the following equation:

Van travel Distance - Car Travel Distance = 0 starting at 1.286 hr.

Let t be the time after 1.286 hours that both vehicles meet/collide.

t*(60 km/h) + t(84 km/h) = 30.86 km

t(60+84) = 30.86 km

t(144 km/h) = 30.86 km

t = 0.2143 hr

Total time until the car and van meet is 1.286 hr + 0.2143 hr for a total of 1.50 hours.

=================

a) Time until both vehicles meet is 1.5 hours after starting at noon. That makes it 1:30PM.

b) Average speed of car is 84 km/h

==============

CHECK

Is the ratio of the distance travelled by the car and the van until they meet in the ratio of 7/5?

Car distance is (1.5 hr)(84 km/h) = 126 km

Van distance is (1.5 hr)(60 km/h) = 90.0 km

Ratio is 126/90 or 1.4

Ratio of 7/5 is 1.4

YES

3 0

2 years ago

Which point is located at (2, -3)?point Hpoint Epoint Gpoint K

Fynjy0 [20]

Answer:

Since i dont have a graph il just tell you this way. on the x axis, go right 2 times, then go down 3 times and that will be your answer.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0

MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

I₂= ∫[-x/(1 + x⁴)] dx

for u= x² → du=x*dx

I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂

then

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C

for y(x=1) = 0

1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C

since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be

1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0

3 years ago