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3 years ago

6

One manufacturer boxes screws so that each box weighs the same no matter what size the screws are. The graph shows the amount of

allowed weight remaining in a given shipment depending on the number of boxes of screws already included.

1 answer:

Rus_ich [418]3 years ago

7 0

I don’t really know but keep trying

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Please help this is due today

natita [175]

Answer:

use this formula

Step-by-step explanation:

I hope this helps ......

5 0

2 years ago

Read 2 more answers

Help pls <br> 8 + 6x = -4

gavmur [86]

Answer:

x=-2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Justin has $7.50 more than Eva, and Emma has $12 less than Justin. Together , they have a total of 63.00. How much money does ea

Setler [38]

Since no value is given to Eva, we'll call her x. Justin, then, is 7.50+x. That leaves Emma, who would be Justin (7.50+x) -12. If we add them all together we get 63. So our equation is

x + (7.50+x) + (7.50+x)-12 = 63

Combine like terms

x+x+x + 7.50+7.50-12 = 63
3x + 3 = 63
3x = 60
x = 20

Since we said Eva is x then Eva has 20 dollars.

Justin is 7.50 + x so 7.50+20=
$27.50

Emma is Justin - 12 so 27.50-12= $15.50

Let's check our answer...

15.50+27.50+20 = 63
63=63

It checks!

4 0

4 years ago

Six times the sum of a number and 7 equals 2

pogonyaev

6(x+7)=2
6(x)=-5
X=-0.83

7 0

3 years ago

A) Let X be a random variable that can assume only positive integer values, and assume its probability function is P(X -n) A/3^n

kramer

Answer:

a) The value of A = 2

b) The value of $B = \dfrac{1}{12}$

Step-by-step explanation:

a)

Given that:

X should be the random variable that assumes only positive integer values.

The probability function; $P[X = n] = \dfrac{A}{3^n}$ for some constant A and n ≥ 1.

Then, let $\sum \limits ^{\infty}_{n =1} P[X =n] = 1$

This implies that:

$A \sum \limits ^{\infty}_{n =1} \dfrac{1}{3^n}= 1$

$A \times \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = 1$

$A \times \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = 1$

$A \times \dfrac{1}{2}=1$

A = 2

Thus, the value of A = 2

b)

Suppose X represents a e constant A (n> 1). Find A.

b) Let X be a continuous random variable that can assume values between 0 and 3

Then, the density function of x is:

$f_x(x) = \left \{ {{B(x^2+1)} \ \ \ 0 \le x \le 3 \ \ \ \atop {0} \ \ \ otherwise} \right.$

where; B is constant.

Then, using the property of the probability density function:

$\int ^3_0 \ B (x^2+1 ) \ dx = 1\\$

Taking the integral, we have:

$B \Big [\dfrac{x^3}{3} +x \Big ]^3_0 = 1$

$B \Big [\dfrac{3^3}{3} +3 \Big ]= 1$

$B \Big [\dfrac{27}{3} +3 \Big ] = 1$

B [ 9 +3 ] = 1

B [ 12 ] = 1

Divide both sides by 12

$B = \dfrac{1}{12}$

5 0

3 years ago