What does > mean. Greater than or less than.

VashaNatasha [74]

Answer: The y is going to be 7

Explanation: Because 5y/35 divide 5 both sides cancel the both 5's and y stays by itself and 35/5 equals 7. 7*5= 35

8 0

2 years ago

We gave a national verbal proficiency test that has a mean score of 500 with a standard deviation of 75. What raw score would co

charle [14.2K]

Answer:

519

Step-by-step explanation:

Let's assume that the scores for the national verbal proficiency test follow a normal distribution.

Mean score (μ) = 500

Standard deviation (σ) = 75

According to a Z-score table, the score for the 60th percentile is roughly z = 0.253

For any score "x", the z-score is given by:

$Z=\frac{X-\mu}{\sigma}$

For z = 2.53:

$0.253=\frac{X-500}{75}\\X=519$

The raw score to the 60th percentile is 519.

4 0

3 years ago

I NEED HELP GUYS PLEASE I CANT FAIL AGAIN A translation was applied to square ABCD to form

klio [65]

B is the right answer

8 0

2 years ago

Read 2 more answers

The equation r(t) = sin(4t)i + cos(4t)j, 0t≥0 describes the motion of a particle moving along the unit circle. Answer the follo

lorasvet [3.4K]

Answer:

a) Particle has a constant speed of 4, b) Velocity and acceleration vector are orthogonal to each other, c) Clockwise, d) False, the particle begin at the point (0,1).

Step-by-step explanation:

a) Let is find first the velocity vector by differentiation:

$\vec v = \frac{dr_{x}}{dt} i + \frac {dr_{y}}{dt} j$

$\vec v = 4\cdot \cos 4t\, i - 4 \cdot \sin 4t \,j$

$\vec v = 4 \cdot (\cos 4t \, i - \sin 4t\,j)$

Where the resultant vector is the product of a unit vector and magnitude of the velocity vector (speed). Velocity vector has a constant speed only if magnitude of unit vector is constant in time. That is:

$\|\vec u \| = 1$

Then,

$\| \vec u \| = \sqrt{\cos^{2} 4t + \sin^{2}4t }$

$\| \vec u \| = \sqrt{1}$

$\|\vec u \| = 1$

Hence, the particle has a constant speed of 4.

b) The acceleration vector is obtained by deriving the velocity vector.

$\vec a = \frac{dv_{x}}{dt} i + \frac {dv_{y}}{dt} j$

$\vec a = 16\cdot (-\sin 4t \,i -\cos 4t \,j)$

Velocity and acceleration are orthogonal to each other only if $\vec v \bullet \vec a = 0$ . Then,

$\vec v \bullet \vec a = 64 \cdot (\cos 4t)\cdot (-\sin 4t) + 64 \cdot (-\sin 4t) \cdot (-\cos 4t)$

$\vec v \bullet \vec a = -64\cdot \sin 4t\cdot \cos 4t + 64 \cdot \sin 4t \cdot \cos 4t$

$\vec v \bullet \vec a = 0$

Which demonstrates the orthogonality between velocity and acceleration vectors.

c) The particle is rotating clockwise as right-hand rule is applied to model vectors in 2 and 3 dimensions, which are associated with positive angles for position vector. That is: $t \geq 0$