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3 years ago

The area of a rectangle is 20 meters. How long is the length if the width is 4 meters?

Mathematics

2 answers:

IgorLugansk [536]3 years ago

7 0

5 is the width
i hope this helps!

miss Akunina [59]3 years ago

4 0

Area = length * Width
20 = length * 4
l = 20/4
l = 5

In short, Your Answer would be 5 meter

Hope this helps!

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You decide to get an estimate on the cost of repairs for the green food truck. The mechanic explains that the cost of the repair

I am Lyosha [343]

Answer:

you save labor : 3562.50

Step-by-step explanation:

You want to calculate for labor.

labor = 3/5 (parts)

labor + parts = $9500 (total costs)

3/5 parts + parts =9500

3/5 parts + 5/5 parts = 9500

8/5 parts=9500

5/8(8/5 parts) = 5/8 (9500)

‘parts = 5937.50

labor=3562.50 (3/5 x 5937.50)

total = 9500

7 0

3 years ago

Graph the image of the given triangle after the transformation with the rule (x, y)→(x, −y) .

Zarrin [17]

Firstly, we will select three corner points

A=(2,3)

B=(6,8)

C=(7,4)

we are given

the transformation with the rule (x, y)→(x, −y)

y--->-y

so, it is reflected about x-axis

so, we will multiply y-value by -1

we get new points as

A=(2,-3)

B=(6,-8)

C=(7,-4)

now, we can locate these points and draw graph

we get

4 0

3 years ago

Read 2 more answers

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

Can anybody help me with this question?

Bogdan [553]

For number 16 we need to write the data as a ratio then convert to a unit rate (amount per 1)...

308/14 = x/1

Cross multiply

14x = 308
x = 308/14
x = 22

So the fuel efficiency is
22 miles per 1 gallon or

22/1

For number 17, since the car was driven at 48 mph, we just have to divide distance driven by speed to get how long it took...

288 miles ÷ 48 mph =

6 hours

6 0

3 years ago

Pls help i will mark brainleist

lara [203]

Answer:

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Step-by-step explanation:

6 0

3 years ago