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2 years ago

Tim's family is visiting Emeryville. They plan to see a movie and then explore the town. The

Mathematics

1 answer:

kap26 [50]2 years ago

5 0

Answer. 4 hours

Step-by-step explanation:

48 + 8 + 8 + 8 + 8

$8 an hour and it $32 =4 hours and since they already using $48 for the movie if the they explore the town for 4 hours and then leave it will $80 in total spent

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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

The temperature, T , of a given mass of gas varies inversely with its volume, V . The temperature of a certain gas with volume o

alekssr [168]

We know that

if the temperature T of a given mass of gas varies inversely with its volume V
then
T=k/V

Step 1
Find the value of k
for T=30º C and V=105 cm³
we have
T=k/V--------> k=T*V--------> k=30*105=3150 °C*cm³

therefore
for V=84 cm³
T=3150/84=37.5 °C

the answer is 37.5 °C

3 0

2 years ago

3. WHAT IS THE END BEHAVIOR?SHOW WORK. - 10x -x+ 2

nekit [7.7K]

Answer:

Step-by-step explanatio

−10x−x+2

=−10x+−x+2

=(−10x+−x)+(2)

=−11x+2

Answer:

=−11x+2

3 0

2 years ago

Area of the figure: A)128 B)136 C)153 C)258

GarryVolchara [31]

Answer: 153

Step-by-step explanation:

area of rectangle= 16 x 3= 48

area of trapezium= 1/2(a+b)h

=1/2(16+5)x 10

=105

105+48= 153 in^2

3 0

2 years ago

What is the domain and range of the set? {(2,2.3),(5,2.3), (7,15)}

aleksandrvk [35]

Answer:

DOMAIN: 2, 5, 7

RANGE: 2.3, 2.3, 15

Step-by-step explanation:

Domain is always the first value of the set, and range is always the second value of the set. Hope it makes sense, and that it helped.

Have a good day! goodluck.

3 0

3 years ago