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3 years ago

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The dimensions of a rectangular garden were 4m by 5m. each dimension was increased b the same amount. the garden then had an are

a of 56m^2. find the dimensions of the new garden.

1 answer:

musickatia [10]3 years ago

4 0

It was increased by 3.
the new dimensions are 7m and 8m. Hope it helps!

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Please answer this correctly

Lynna [10]

Answer:

557

Step-by-step explanation:

l x w

13x24

13x7

22x7

557

4 0

3 years ago

Does second-hand smoke increase the risk of a low birthweight? A baby is considered have low birthweight if he/she weighs less t

marishachu [46]

Answer:

Option 2) Null hypothesis: p = 0.078 , Alternate hypothesis: p > 0.078

Step-by-step explanation:

We are given the following in the question:

According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birth weight.

Sample size, n = 1200

p = 7.8% = 0.078

We have to carry a hypothesis test whether national percentage is higher than 7.8% or not.

Thus, we can design the null and the alternate hypothesis

$H_{0}: p = 0.078\H_A: p > 0.078$

Thus, the correct answer is:

Option 2) Null hypothesis: p = 0.078 , Alternate hypothesis: p > 0.078

5 0

3 years ago

Please Help Will give brainliest, 5 stars and thanks!

mart [117]

Okay so 0.08L=80mL. 8,000mL=8L. 0.8L=800 mL. And 80mL=0.08L. Hope this all helps

6 0

3 years ago

Two lighthouses flash their lights every 20 and 30 seconds respectively. given that they flash together at 8 p.m., at what time

VikaD [51]

Do you remember when you were learning about
Least Common Multiple (LCM) ? And you were
wondering what on Earth you would ever need it for ?

THIS IS IT !

The amount of time after 8 PM when the lights will flash
together again is the LCM of 20 seconds and 30 seconds.
Do you remember how to find it ?

... When 20 is prime-factored, you have 2 · 2 · 5 .

... When 30 is prime-factored, you have 2 · 3 · 5 .

... So the Least Common Multiple of 20 and 30 is 2 · 2 · 3 · 5 = 60 .

The lighthouses will flash together again in 60 seconds after 8 PM.
That'll be 8:01 PM.

6 0

3 years ago

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

3 years ago