Please help please show how you got the answer please I'm so confused on how you get the answer

For know the minimum value to be eaten daily to provide the requirement of both vitamins, is necessary to calculate the minimum value for every vitamin, so:

vitamin E: We are going to use the rule of three in which 1 gram have 7% of the minimum daily requirement, then how many grams are going to be be the 100% of the daily requirement:

1 gram --------------7%

X grams -----------100%

$X=\frac{1 gram* 100}{ 7 } = 14.285 grams$

So, it is necessary approximately 15 grams of healthy food to complete th 100% of the minimum daily requirement of vitamin E.

vitamin A: We are going to use the rule of three in which 1 gram have 3% of the minimum daily requirement, then how many grams are going to be be the 100% of the daily requirement:

1 gram --------------3%

X grams -----------100%

$X=\frac{1 gram* 100}{ 3 } = 33.33 grams$

So, it is necessary approximately 34 grams of healthy food to complete th 100% of the minimum daily requirement of vitamin E.

Finally for satisfy with both minimum daily requirement, we need to eat at least 34 grams of healthy food, because it is the maximum between the two X values.

3 0

3 years ago

When doing blood testing for a viral infection, the procedure can be made more efficient and less expensive by combining partial

Serjik [45]

Answer:

0.1694 = 16.94% probability of a positive result for three samples combined into one mixture.

Step-by-step explanation:

For each test, there are only two possible outcomes. Either it is positive, or it is negative. The probability of a test being positive or negative is independent of any other test, which means that the binomial probability distribution is used 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.

The probability of an individual blood sample testing positive for the virus is 0.06.

This means that $p = 0.06$

If samples from three people are combined and the mixture tests negative, we know that all three individual samples are negative. Find the probability of a positive result for three samples combined into one mixture.

It will be positive if at least one of the tests is positive, that is:

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

In which

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

$P(X = 0) = C_{3,0}.(0.06)^{0}.(0.94)^{3} = 0.8306$

Then

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8306 = 0.1694$

0.1694 = 16.94% probability of a positive result for three samples combined into one mixture.

8 0

2 years ago

Bill plans to buy school supplies and the following equation represents what he bought.

iren2701 [21]

Answer:

1. x = 8

2. y = 1/2

Step-by-step explanation:

1. Plug the function

5x + 20(0.50) = 50

= 5x + 10 = 50

Subtract 10 from both side

= 5x = 40

Divide 5 on both sides

= x = 8

2. Plug in the function

5(8) + 20y = 50

= 40 + 20y = 50

Subtract 40 on both sides

= 20y = 10

Divide 20 both sides

= y = 10/20 | 1/2

6 0

2 years ago

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