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Debora [2.8K]
3 years ago
6

There are 18 oranges in box A and 36 in box B. What is the ratio of the oranges in Box A to Box B?

Mathematics
2 answers:
Gwar [14]3 years ago
5 0

Answer:

hello there the answer is 36:18

Step-by-step explanation:

because it dosent matter which number is bigger it matters about the oder they ask to put it in and in this case a to b =36to18


xxTIMURxx [149]3 years ago
4 0

Hey there!

\boxed{BoxA: 18}

\boxed{BoxB:36}

You could have your ratio to look like this: 18:36 (most likely going to be your answer if you are looking for the ratio of the number)

<h2>BUT if you're trying to solve your ratio then....</h2>

Both terms go into 18, 9, and 2

\frac{18}{2} = 9

\frac{36}{2} = 18

\boxed{9:18}

\frac{18}{9} = 2

\frac{36}{9} = 4

\boxed{2:4}

\frac{18}{18} = 1

\frac{36}{18} = 2

\boxed{1:2}

Answer: \boxed{36:18} or \boxed{18:36}

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

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Answer:

Let's calculate:

22-5

• Remove minus in the exponent and invert the base:

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3 years ago
The blood platelet counts of a group of women have a​ bell-shaped distribution with a mean of 247.9 and a standard deviation of
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Answer:

A) Approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) approximate percentage of women with platelet counts between 53.8 and 442.0 = 99.7%

Step-by-step explanation:

We are given;

mean;μ = 247.9

standard deviation;σ = 64.7

A) We want to find the approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3.

Now, from the image attached, we can see that from the empirical curve, the probability of 1 standard deviation from the mean is (34% + 34%) = 68 %.

While probability of 2 standard deviations from the mean is (13.5% + 34% + 34% + 13.5%) = 95%

Thus, approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) Now, we want to find the approximate percentage of women with platelet counts between 53.8 and 442.0.

53.8 and 442.0 represents 3 standard deviations from the mean.

Let's confirm that.

Since mean;μ = 247.9

standard deviation;σ = 64.7 ;

μ = 247.9

σ = 64.7

μ + 3σ = 247.9 + 3(64.7) = 442

Also;

μ - 3σ = 247.9 - 3(64.7) = 53.8

Again from the empirical curve attached, we cans that at 3 standard deviations from the mean, we have a percentage probability of;

(2.35% + 13.5% + 34% + 34% + 13.5% + 2.35%) = 99.7%

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4 0
3 years ago
Read 2 more answers
You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r
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Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

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Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. 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.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that p = 0.15

Assume you obtain a random sample of 9 individuals from this population:

This means that n = 9

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

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

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

So

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

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679

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P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0
2 years ago
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