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3 years ago

Sarah received a gift in a rectangular box with a volume of 240 in3. If the base of the box was 20 in2, how tall was the box?

Mathematics

1 answer:

Lunna [17]3 years ago

8 0

Who was this leader of lndia often associated with acts of nonviolent resistance

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What is x in -2/3x - 11 = -7

mafiozo [28]

-2/3x -11=-7
-2/3x = 11 - 7

-2/3x = 4

x = 4 : (-2/3)

x = 4 * (-3/2)

x = 4 *(-3) : 2

x = -12 : 2

x = -6

7 0

4 years ago

(y=x-4 (y=-2x+5 ................................................................

harkovskaia [24]

I don’t know the answer

4 0

3 years ago

A poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life. What is the proba

agasfer [191]

Answer:

The probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.

Step-by-step explanation:

Let X = number of Americans who believe that statistics teachers know the true meaning of life.

The probability of the random variable X is,

P (X) = 0.574

The event of a person not believing that statistics teachers know the true meaning of life is the complement of the event X.

The probability of the complement of an event, E is the probability of that event not happening.

$P(E^{c})=1-P(E)$

Compute the value of $P(X^{c})$ as follows:

$P(X^{c})=1-P(X)\\=1-0.574\\=0.426$

Thus, the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.

4 0

3 years ago

Read 2 more answers

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 1)(t - 3)3, t > 0. /p>For what va

Zepler [3.9K]

Given the position function, the velocity function is obtained by taking the derivative:

$s(t)=(t+1)(t-3)^3\implies v(t)=(t-3)^3+3(t+1)(t-3)^2$

The velocity is increasing its own derivative is positive, so we also have to find the acceleration by taking another derivative:

$a(t)=4(t-3)^3+3(t-3)^2+6(t+1)(t-3)$

To find when $a(t)>0$ , we first need to know where $a(t)=0$ :

$4(t-3)^3+3(t-3)^2+6(t+1)(t-3)=(t-3)\bigg(4(t-3)^2+3(t-3)+6(t+1)\bigg)=(t-3)(4t^2-15t+33)=0$

The quadratic factor is always positive (its discriminant is negative), which leaves one solution at $t=3$ . To either side of $t=3$ , we have, for instance,

$a(2)=-12$

$a(4)=36>0$

which indicates that $v(t)$ is increasing for $t>3$ , making the answer A.

5 0

3 years ago

PLZ HELP ME What shape is this cross-section?

Irina18 [472]

Answer:

i'm not sure if i'm right but think is cube

6 0

4 years ago