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3 years ago

If you have 1 /18 profit out of 75 boxes how many more to make 180 dollars

2 answers:

7 0

Assuming that means you make $0.06 per box (1/18 dollars per box) then you've made $4.17 off of the 75 boxes. 180-4.17 = 175.83 / .06 = 3165 more boxes.

You'd need to sell 3,165 more boxes.

IrinaK [193]3 years ago

6 0

Computing the 75% increase: .... Example 18 ... If you were to put $683.7 billion in $100 bills, and count out 1 per second, it would ... Rectangular Box Cylinder.

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Please help with 4. And 5

Hoochie [10]

More info please thanks

6 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

8. Simplify and solve this equation: 4m + 9 + 5m – 12 = 42. A. m = 5 B. m = –5 C. m = 4 D. m = -4

STatiana [176]

4m+9+5m-12=42 9m-3=42 Divide bothsides by 9. 9m/9 -3/9=-0.3=-4 m=-4

4 0

3 years ago

Read 2 more answers

Menos 10 * 12 * -4 + 40 * -2 * 6 - 2 por favor

SVEN [57.7K]

Answer:

-2

Step-by-step explanation:

-10 x 12 x -4 + 40 x -2 x 6 - 2

-120 x -4 - 80 x 6 - 2

480 - 480 - 2

0 - 2

-2

8 0

3 years ago

Proper angle for a ladder is about 75° from the ground. Suppose you have a 10 foot ladder. How far from the house should you pla

Alja [10]

Answer:

The base of the ladder is 2.58 m.

Step-by-step explanation:

Given that,

The angle of elevation for a ladder from the ground is 75°.

The length of a ladder, H = 10 foot

We need to find the distance from the house should you place the base of the latter. Let the base of the ladder is b. Using trigonometry,

$\cos\theta=\dfrac{B}{H}\\\\B=H\times \cos\theta\\\\B=10\times \cos(75)\\\\B=2.58\ m$

So, the base of the ladder is 2.58 m.

8 0

3 years ago