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2 years ago

The difference of 5 times y and 3

1 answer:

3 0

The way to find the difference between 5y - 3 you have to first do the inverse operation of multiplication which is division.

5y - 3
__ __
5 5

y - 0.6

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1. A student lifts a box of books that weighs 200 N. The box is lifted 2.5 m. How much work

Angelina_Jolie [31]

The student does a work of 500 Newton meter on the box.

Step-by-step explanation:

Given,

Weight of box of books = force = 200 N

Distance the box is lifted = 2.5 m

Work done = Force * distance

Work done = $200*2.5$

Work done = 500 Newton-meter

Newton meter is same as Joule.

Work done = 500 Joules

The student does a work of 500 Newton meter on the box.

Keywords: multiplication, work done

Learn more about multiplication at:

#LearnwithBrainly

7 0

3 years ago

How many times does 1/4 go into 3/8?

Luba_88 [7]

3/8 divided by 1/4

hope this helped

5 0

2 years ago

Read 2 more answers

Sketch the graph of this linear equation<br>4y = -7x + 1

jonny [76]

Answer:

use desmos to graph it

.....

6 0

3 years ago

50 Points and brainliest, simple data

Masja [62]

14 | 7 | 21
6 | 3 | 9
20 |10| 30

i hope you understand how im trying to put i

7 0

2 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