Ask question

Ask question

All categories

3 years ago

14

Let y = safe load in pounds and x = length in feet of a horizontal beam. A constant of proportionality k exists such that y=k/x

. y varies inversely as x. Determine the constant k for a beam with y = 2,000 pounds and x = 15 feet. a. 133.3 b. 3,000 c. 30,000

1 answer:

Contact [7]3 years ago

7 0

Answer:

Option c. 30,000

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent an inverse variation if it can be expressed in the form $y*x=k$ or $y=k/x$

Let

y ----> safe load in pounds

x ----> length in feet of a horizontal beam

we have

For $x=15\ feet, y=2,000\ pounds$

Remember that

$k=y*x$

substitute

$k=2,000*15=30,000$

You might be interested in

Which pair of triangles is similar?

irina [24]

I believe the answer is the fourth option....because 13 × 3 is 39 5 × 3 is 15 and 12 × 3 is 36

6 0

3 years ago

Read 2 more answers

Danny wants to buy a truck in 4 years. He is going to put away $2,500.00 into his savings account that will pay him 6.75% intere

aniked [119]

Answer:

b $3,272.43

Step-by-step explanation:

A = p(1+r/n)^nt

Where

A= future value

P= principal = $2500

r= interest rate = 6.75% = 0.0675

n = number of periods = 12

t = time = 4 years

A = p(1+r/n)^nt

= 2500(1+0.0675/12)^12*4

= 2500(1+0.005625)^48

= 2500(1.005625)^48

= 2500(1.3089737859257)

= 3272.4344648144

Approximately

A= $3272.43

He will have $3272.43 to give as down payment in 4 years

4 0

3 years ago

Choose the numbers to complete the statement about the expression.

mrs_skeptik [129]

Answer:

1 + and a half = 1 in a half

Step-by-step explanation:

The numerator in the first fraction is closest to

10, so the fraction is nearest to 1.

The numerator in the second fraction is closest to 3, so the fraction is nearest to one-half.

The value of the expression can be estimated as 1 + one-half = 1 and one-half.

5 0

2 years ago

Help :( I need this :)

valina [46]

Answer:

you can put in any of the numbers that are already listed in the x values. You can put 5, 14, -1, and 23 because something can't be a function if x values repeat.

8 0

3 years ago

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.

6 0

3 years ago