Ask question

Ask question

All categories

3 years ago

15

How can you show that two objects are proportional with a graph?

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

So this is our table:

-----------------
x-values | a b c d e f |
-----------------
y-values | g h i j k l |
-----------------

You can know that the values in the table are proportional if:

$\frac{a}{g}=\frac{b}{h} =\frac{c}{i}=\frac{d}{j}=\frac{e}{k}=\frac{f}{l}$

Hope that helps :)

You might be interested in

Find the 35th term in the arithmetic sequence: -10,-14,-18,-22,..

ivanzaharov [21]

The 35th term is -146 :)

3 0

3 years ago

What is 1/2x+1.5-2-7 simplified

Nady [450]

The answer is x/2-7.5 !!

6 0

3 years ago

The distance from first base to third base is 1.4times the distance from home plate to third base, which is k feet. REPRESENT TH

geniusboy [140]

Answer:

1.4 times k

Step-by-step explanation:

If k feet is the distance from home plate to third base, and the distance from first base to third base is 1.4 times the previous value, then

distance from first base to third base = 1.4 * k or 1.4 times k

If, for example, k=90 feet, then from first to third base there are

1.4*90=126 feet

4 0

3 years ago

For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

6 0

3 years ago

Please, I need help with this! :(

Serggg [28]

Answer:

Step-by-step explanation:

7 0

3 years ago