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2 years ago

Directions: Select ALL the correct answers.

Mathematics

1 answer:

dybincka [34]2 years ago

7 0

Answer:

b, that each player loaded onto the truck is

proportional to the number of minutes. m. spent loading boxes.

• Cory loaded twice as many boxes as Alonzo.

The equation that represents the number of boxes Cory loaded is

b = 2m.

Step-by-step explanation:

hope this helps

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I sailed west to Hawaii against the wind at the speed of 12 knots/hour. When I returned, I sailed the same route with the wind a

expeople1 [14]

Answer:

Step-by-step explanation:

total time is 5 hr and total distance is 72 nautical miles

7 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

2 years ago

Solve: 10(1 + 3y) = 20 <br> What does y equal?<br> Y =

marin [14]

Answer:

y = 1/3 ≈ 0.3

hope this helps! :)

6 0

2 years ago

Help me with this please no links

Nostrana [21]

The last option is the answer

8 0

2 years ago

Question 5

Stells [14]

Answer:

7**6

Step-by-step explanation:

The expression 7^6 ; can be interpreted as 7 raised to the power of 6 ; which is (7 * 7 * 7 * 7 * 7 * 7). To execute this expression using a computer or basic mathematical operation on a computer we use the power symbol which the computer understand this the double multiplication symbol (**)

Hence, 7^6 = (7**6)

7 0

2 years ago