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cupoosta [38]
2 years ago
5

Mr Davis is creating a spice mixture for a recipe.2/5 of the spice mixture was oregano 1/3 of the spice mixture was basil the re

maining spice mixture was chili powder what fraction of the total amount of spice mixture was oregano and basil
Mathematics
1 answer:
attashe74 [19]2 years ago
7 0

Answer: \frac{11}{15}

Step-by-step explanation:

You know that:

- 2/5 of the spice mixture was oregano.

- 1/3 of the spice mixture was basil.

Then, to find the fraction of the total amount of spice mixture that was oregano and basil, you must add both fractions, as following:

- Find the least common multiply of the denominators:

LCM=5*3=15

- Divide the LCM by each original denominator and multiply the result by each numerator.

- Make the addition.

Then, the result is:

\frac{(2*3)+(1*5)}{15}=\frac{6+5}{15}=\frac{11}{15}

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The answer to the example is -8 so the answer is B. negative
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A stack of cards consists of seven red and four blue cards. A second stack of cards consists of eleven red cards. A stack is sel
Ann [662]

Answer:

0.1733 = 17.33% probability the first stack was selected.

Step-by-step explanation:

To solve this question, it is needed to understand conditional probability, and the hypergeometric distribution.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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)!}

Conditional probability:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

Probability of all red cards for the first stack:

For this, we use the hypergeometric distribution, as the cards all chosen without replacement.

7 + 4 = 11 cards, which means that N = 11.

7 red, which means that k = 7

3 are chosen, which means that n = 3

We want all red, so we find P(X = 3).

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 3) = h(3,11,3,7) = \frac{C_{7,3}*C_{4,0}}{C_{11,3}} = 0.2121

Conditional probability:

Event A: All red

Event B: From the first stack.

Probability of all red cards:

0.2121 of 50%(first stack)

1 of 50%(second stack). So

P(A) = 0.2121*0.5 + 1*0.5 = 0.60605

Probability of all red cards and from the first stack:

0.21 of 0.5. So

P(A \cap B) = 0.21*0.5 = 0.105

What is the probability the first stack was selected?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.105}{0.60605} = 0.1733

0.1733 = 17.33% probability the first stack was selected.

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The answer would be 20 apples. 
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Answer:

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2r > - 8.6 + 4.4

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The least integer is <em>- 2</em>

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