A dairy farmer decides to sell three fifth of her 500 cows. How many cows will she be left with after the deal is complete?

1 answer:

3 0

Answer is 200 cows
Step by step
She is selling 3/5 of 500
Multiply across 3/5 x 500
1500/5 divide to get 300 sold
Now 500-300 sold = 200 cows left
Problem solved

You might be interested in

QUESTION 21 ✓ Round 5 1/3 to the nearest whole number?

wariber [46]

Answer:

Step-by-step explanation:

1/3 is only 0.33 and 0.33 isnt higher than 0.50, so you cant round anything. 5 is the answer.

5 0

2 years ago

Read 2 more answers

bobbie wrote y+6=15. than she wrote (y+6) divided by 3=15. Explain why the second equation is not equivalent to the first. what

irina1246 [14]

If y+6=15, then y=9. (9+6)divided by 3 would be 5, not 15.

4 0

3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$