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2 years ago

7

Carol is going to be away on vacation for a couple of days. She does not know how long she will be gone. Carol, however, knows t

hat $250 will be enough for the caretaker to feed the cats, Rover and Bobo, while she is away.
How many days can the caretaker feed the cats before Carol returns? Explain your reasoning critically.

1 answer:

Sliva [168]2 years ago

6 0

Is there more to the question? Like how much the cats eat a day or the price of the food?

You might be interested in

There are two coins, one is fair and one is biased. The biased coin has a probability of landing on heads equal to 4/5. One of t

den301095 [7]

Answer:

72.4%

Step-by-step explanation:

The probability of A occurring given that B occurs = the probability of both A and B / the probability of B

P(A|B) = P(A∩B) / P(B)

This can be rearranged as:

P(A∩B) = P(B) P(A|B)

In this case:

A = biased coin is chosen

~A = fair coin is chosen

B = 4 heads then 1 tail

First, let's find P(A∩B).

P(A∩B) = P(B) P(A|B)

P(A∩B) = ½ × ₅C₄ (⅘)⁴ (⅕)¹

P(A∩B) = 0.2048

Next, find P(~A∩B).

P(~A∩B) = P(B) P(~A|B)

P(~A∩B) = ½ × ₅C₄ (½)⁴ (½)¹

P(~A∩B) = 0.078125

Therefore, the probability that the coin is biased is:

P = P(A∩B) / (P(A∩B) + P(~A∩B))

P = 0.2048 / (0.2048 + 0.078125)

P = 0.723866749

The probability is approximately 72.4%.

3 0

3 years ago

Anyone good at statistics? Please help me

Lyrx [107]

I am good at it i can help u

4 0

3 years ago

Any one can help with this Venn Diagram question?.Please note that there is an another language, ignore except if you can unders

zloy xaker [14]

Answer:

(i) 30

(ii) 8

Step-by-step explanation:

Start by drawing 3 rings for the Venn diagram. Each ring represents either Audi, BMW, or Citroën.

Next, add the information provided in the problem statement. We know that 20 choose BMW only. Another 20 choose Citroën only.

We know the number who choose all 3 is the same as the number that choose BMW and Citroën, but not Audi. We'll call this number x.

Similarly, we know that the number who choose all 3 is half of those choose Audi and BMW, but not Citroën. So we'll say that number is 2x.

We know the total number of Audi votes is 48, the total number of BMW votes is 36, and the total number of Citroën votes is 34.

Finally, we know the total number of voters is 100, including those who voted for no favorites (we'll call that number n).

We know that BMW got 36 votes, so:

20 + x + x + 2x = 36

20 + 4x = 36

4x = 16

x = 4

We know that Citroën got 34 votes, so:

20 + x + x + z = 34

20 + 4 + 4 + z = 34

28 + z = 34

z = 6

We know that Audi got 48 votes:

y + 2x + x + z = 48

y + 2(4) + 4 + 6 = 48

y + 18 = 48

y = 30

Finally, we know that the total number of voters is 100.

20 + 20 + x + x + 2x + y + z + n = 100

20 + 20 + 4 + 4 + 2(4) + 30 + 6 + n = 100

92 + n = 100

n = 8

8 0

3 years ago

Which linear inequality is represented by the graph?

Fantom [35]

Answer:

y $\leq$ 1/2x + 2 :)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

So

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago