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3 years ago

Help Please in the file. (d)

Mathematics

1 answer:

hichkok12 [17]3 years ago

8 0

The crane cost 800,000.....because u r basically subbing in 0 for t , because the crane has not been used.

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Find the total cost of 5 bales of hay at $2.75 per bale and 2 bags of potatoes at $2.90 per bag.

Elan Coil [88]

Answer: $19.55

Step-by-step explanation:

2.75 * 5 = 13.75.
2.90 * 2 = 5.80
13.75 + 5.80 = 19.55

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4 years ago

An airplane travels about 1500 km in a straight line between toronto and winnipeg. however, the plane must climb to 10 000 m dur

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3 years ago

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Marta_Voda [28]

Answer:

x=2

y=-3

Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the average rate of change of y with respect to x over the interval [-2, 6] for the function y = 5x 2?

PolarNik [594]

The rate of change for the points (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)
given
y=5x^2 and find the slope between x=-2 and x=6

y=6(-2)^2=6(4)=24
y=6(6)^2=6(36)=216

(-2,24)
(6,216)

slope=(216-24)/(6-(-2))=(192)/(6+2)=192/8=96/8=48/4=24/2=12
the rate of change is 12

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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