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3 years ago

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In a Gallup Poll in 2003, 590 of 1,116 randomly selected adults aged 18 and older said they had too little time for relaxing or

doing nothing. Identify the upper level of a 95% confidence interval of the true proportion of adults who say they have too little time for relaxing.

1 answer:

White raven [17]3 years ago

8 0

Answer:

hope this helps

Step-by-step explanation:

what do you have to do

You might be interested in

Select all values equivalent to -10/7<br> -10/-7<br> -3 1/7<br> 1 3/7<br> - -10/-7<br> -1 3/7

Kitty [74]

We know that

case 1) -10/-7-----> 10/7-------> is not <span>equivalent to -10/7

case 2) </span>-3 1/7----> (-3*7+1)/7----> -20/7 ------> is not equivalent to -10/7

case 3) 1 3/7-----> (1*7+3)/7----> 10/7 ------> is not equivalent to -10/7

case 4) - -10/-7---> +10/-7----> -10/7------> is equivalent to -10/7

case 5) -1 3/7----> (-1*7+3)/7----> -4/7 ------> is not equivalent to -10/7

the answer is
- -10/-7

5 0

3 years ago

Un escalador sube 225m de altura. El asenso lo hizo en 3 etapas. En la primera subio un quinto en la, en la tercera un cuarto. ¿

kiruha [24]

Answer:

Segunda etapa= 123.75 metros

Step-by-step explanation:

Altura total= 225 metros

<u>En la primera estapa subió el 20% (un quinto):</u>

Primera etapa= 225*0.2= 45 metros

<u>En la tercera etapa subió 25% (un cuarto):</u>

Tercera etapa= 225*0.250 56.25 metros

Ahora debemos determinar cuánto subió en la segunda etapa:

Segunda etapa= altura total - total subido

Segunda etapa= 225 - (45 + 56.25)

Segunda etapa= 123.75 metros

6 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

The value of a and b

Jlenok [28]

Answer:

Both a and b would both equal 45 degrees

Step-by-step explanation:

To find this, we need to note that a and 135 create a straight line. Since a straight line has 180 degrees, we can create an equation to solve for a.

135 + a = 180

a = 45

Now that we know a is equal to 45, we can tell that b is also equal to that amount. This is because two parallel lines cut by a transversal creates the same angle.

3 0

3 years ago

Which of the following is the solution to the system of equations shown below?

antiseptic1488 [7]

Answer:

x = -2, x = 1

Step-by-step explanation:

mark as brainliest it is right

3 0

3 years ago