Each of 18 horses eat 2/3 of a bales of hay how many bales do 18 horses eat

Mathematics

2 answers:

Anton [14]3 years ago

6 0

18 * 2/3 = 12.

The 18 horses eat 12 bales of hay in total.

frutty [35]3 years ago

5 0

Simply put, 18 x 2/3 = 36/3 = 12

The horses eat 12 bales

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What Should i write down basically i just need the answers and let me copy and paste

Anastaziya [24]

Answer:

✔️Ratio of weight to mass in simplest form: 49/5 or 49 : 5

✔️x = 110

Step-by-step explanation:

Since the ratio of weight to mass is constant, therefore we can use any pair of values from the table to find the ratio that represents the table. Let's pick the first row:

(20, 196)

Ratio of weight to mass = 196 : 20

= 196/20

= 49/5

Or 49 : 5

✔️Let's use this ratio/proportion to find x in the table

Ratio of weight to mass = 49:5

Thus:

1078/x = 49/5

Cross multiply

x*49 = 1078*5

49x = 5,390

x = 5,390/49

x = 110

5 0

3 years ago

We have n = 100 many random variables xi 's, where the xi 's are independent and identically distributed bernoulli random variab

Alex777 [14]

Recall that for a random variable $X$ following a Bernoulli distribution $\mathrm{Ber}(p)$ , we have the moment-generating function (MGF)

$M_X(t)=(1-p+pe^t)$

and also recall that the MGF of a sum of i.i.d. random variables is the product of the MGFs of each distribution:

$M_{X_1+\cdots+X_n}(t)=M_{X_1}(t)\times\cdots\times M_{X_n}(t)$

So for a sum of Bernoulli-distributed i.i.d. random variables $X_i$ , we have

$M_{\sum\limits_{i=1}^nX_i}(t)=\displaystyle\prod_{i=1}^n(1-p+pe^t)=(1-p+pe^t)^n$

which is the MGF of the binomial distribution $\mathcal B(n,p)$ . (Indeed, the Bernoulli distribution is identical to the binomial distribution when $n=1$ .)