Ask question

Ask question

All categories

3 years ago

6

In a lab, a substance was cooled by 24 C over a period of 6 hours at a constant rate. What was the change in temperature each ho

ur?

1 answer:

Karo-lina-s [1.5K]3 years ago

6 0

Step-by-step explanation:

Decrease of 24°C in 6 hours

=> -24°C / 6 hours

=> -4°C / hour

You might be interested in

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

Nishi invests £3200 at 6% interest per year. Work out how much she will have altogether after: 2 years

artcher [175]

SI = P×R×T/100

(3200×6×2)/100

= 384

Now 3200 + 384

= Euro 3584

4 0

3 years ago

If one side of a rectangle is increased by 12 and an adjacent side is decreased by 3 a rectangle is formed whose perimeter is 64

Dima020 [189]

The length of the side of the original square is 7cm

8 0

3 years ago

Consider the linear function that is represented by the equation y = negative 10 x + 6 and the linear function that is represent

Vilka [71]

Answer: The function that is represented by the equation y = negative 10 x + 6 has a steeper slope and a greater y-intercept.

Step-by-step explanation: I took the unit test on edge hoped this helped <3

8 0

4 years ago

Read 2 more answers

What is the value of x?

Papessa [141]

The value of x is 2 because we can compare like terms

5 0

3 years ago