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Vlad [161]
3 years ago
12

A field of grass feeds 4 elephants for 4 days. How many days would the same field feed 16 elephants ?

Mathematics
2 answers:
Marta_Voda [28]3 years ago
8 0

Answer:

1 day  

Step-by-step explanation:

16 is 4 times more elephant than 4.

So, divide 4 by 4

1 day

sergejj [24]3 years ago
3 0

Answer:

1 day

Step-by-step explanation:

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3/4k + 3/8k = 1/2<br> solve for k please
vlabodo [156]

Answer:

K = 4/9

Step-by-step explanation:

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3 years ago
Jenny thinks that the expression below is equal to x²-4. If you agree, show that she is correct. If you disagree, show that she
sergiy2304 [10]

Answer:

Jenny is wrong the correct answer is x²+ 4 - 4 x.

Step-by-step explanation:

Given that

(x-2)³∕x-2

(x-2)³∕(x-2) =(x-2) (x-2)²∕(x-2)

(x-2)³∕(x-2) = (x-2)²

As we know that

(a-b)² = a²+b² - 2 ab

So

(x-2)² = x²+ 2² - 2 ˣ 2 ˣ x =  x²+ 4 - 4 x

(x-2)² =  x²+ 4 - 4 x

It means that

(x-2)³∕(x-2) = (x-2)² =  x²+ 4 - 4 x

So Jenny is wrong the correct answer is x²+ 4 - 4 x.

8 0
3 years ago
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7 0
3 years ago
Read 2 more answers
WILL MARK BRAINLEIST PLEASE HELP HURRY
Zina [86]

Answer:

Hello, Your answer will be D.

Step-by-step explanation:

The data is symmetric for Shop A but not for Shop B ( note the values 10 and 11 for Shop B which are a lot lower than the other values).

Mean for Shop A and Median for Shop B. <u><em>Have a Great Day!</em></u>

7 0
4 years ago
If y has moment-generating function m(t) = e 6(e t −1) , what is p(|y − µ| ≤ 2σ)?
Lostsunrise [7]
Since \mathrm M_Y(t)=e^{6(e^t-1}, we know that Y follows a Poisson distribution with parameter \lambda=6.

Now assuming \mu,\sigma denote the mean and standard deviation of Y, respectively, then we know right away that \mu=6 and \sigma=\sqrt6.

So,

\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028
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4 years ago
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