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steposvetlana [31]

3 years ago

12

A box is to be constructed with a rectangular base and a height of h cm. The length, l, of the box is 10 cm, and the width, w, i

s twice the height. Which quadratic equation best models the volume of the box? V = lwh V = 20h V = 20h2 V = 100h V = 100h2.

1 answer:

Andru [333]3 years ago

4 0

Answer:

20h^2

Step-by-step explanation:

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Kim picked 250 pounds of pears yesterday and 287 pounds of pears today. What is the percentage increase of

xxMikexx [17]

Given:

Kim picked 250 pounds of pears yesterday and 287 pounds of pears today.

To find:

The percentage increase of amount of pears picked by Kim.

Solution:

We know that,

$\%\text{ increase}=\dfrac{\text{New value - Initial value}}{\text{Initial value}}\times 100$

$\%\text{ increase}=\dfrac{287-250}{250}\times 100$

$\%\text{ increase}=\dfrac{37}{25}\times 10$

$\%\text{ increase}=\dfrac{370}{25}$

$\%\text{ increase}=14.8$

Therefore, the percentage increase of amount of pears picked by Kim is 14.8%.

7 0

3 years ago

Seven is reading a book while on vacation. So far, he has read 1/6 of the book. He wants to finish 3/6 of the book by the end of

SOVA2 [1]

Answer:

She has to read 2/6 more by the end of her vacation if she wants to finish 3/6 (1/2) of the book.

Step-by-step explanation:

If this wasn't the answer you were looking for please tell me so I can redo it.

4 0

3 years ago

Read 2 more answers

) One year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million. Suppose a samp

Nutka1998 [239]

Answer:

Probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

Step-by-step explanation:

We are given that one year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million.

Suppose a sample of 400 major league players was taken.

<em>Let </em> $\bar X$ <em> = sample average salary</em>

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = mean salary = $1.5 million

$\sigma$ = standard deviation = $0.9 million

n = sample of players = 400

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the average salary of the 400 players exceeded $1.1 million is given by = P( $\bar X$ > $1.1 million)

P( $\bar X$ > $1.1 million) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ > $\frac{ 1.1-1.5}{{\frac{0.9}{\sqrt{400} } }} }$ ) = P(Z > -8.89) = P(Z < 8.89)

<em>Now, in the z table the maximum value of which probability area is given is for critical value of x = 4.40 as 0.99999. So, we can assume that the above probability also has an area of 0.99999 or nearly close to 1.</em>

Therefore, probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

4 0

3 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

3 years ago

Y=3/2x+2 <br> 5x-y=5<br> At what point do they meet?

RSB [31]

(2,5) ...........................................................................................

7 0

3 years ago