Please help! i might be wrong so please correct me. thanks for the help i will reward points!

the gravitational attraction between two masses varies inversely as the square of the distance between them. if the attraction f

Marat540 [252]

Answer:

33 1/3 lb

Step-by-step explanation:

When the distance between them goes from 8 ft to 12 ft, it is 1.5 times what it was. Then the force will be multiplied by the inverse of the square of this:

(75 lb)×1/(1.5²) = 75/2.25 lb = 33 1/3 lb

At 12 feet apart, the attraction force is 33 1/3 pounds.

8 0

4 years ago

A test consists of 10 true or false questions. To pass the test a student must answer at least eight questions correctly. If the

forsale [732]

<h3>The probability of student passing the quiz with at least 50% of the questions correct is 0.05457.</h3>

Step-by-step explanation:

Here, the total number of T/F question = 10

The minimum answers needed correctly answered = 8

So, student needs to answer at least 8 questions correctly.

Here, the possibility of answering a question correctly = $(\frac{1}{2})$ = p = 0.5

Also, the possibility of answering a question wrong = $(\frac{1}{2})$ = q = 0.5

Now, to pass he needs to answer 8 or more ( 8 , 9 or 10) answers correctly.

P(answering 8 correct answer) = $^{10}C_8(p)^8(q)^2 = ^{10}C_8(0.5)^8(0.5)^2 = 0.0439$

P(answering 9 correct answer) = $^{10}C_9(p)^9(q)^1 = ^{10}C_9(0.5)^9(0.5)^1 = 0.0097$

P(answering 10 correct answer) = $^{10}C_{10}(p)^{10}(q)^0 = ^{10}C_{10}(0.5)^{10}(0.5)^0 = 0.00097$

So, the total Probability = P(8) + P(9) + P(10)

= (0.0439) + (0.0097) + (0.00097)

= 0.05457

Hence, the probability that the student passes the quiz with at least 8 of the questions correct is 0.05457.

6 0

4 years ago

What is the Value Of X<br> 2<br> --- X = 4<br> 3

SashulF [63]

2
— x =4
3

3* 2/3 x = 4*3

2x=12
_____

2x = 2

X= 6

3 0

3 years ago

How to do percent equations easily

Alexandra [31]

Make the percent into a decimal.

6 0

3 years ago

t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0

4 years ago