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ankoles [38]
3 years ago
7

One serving of the food product contains 1g total saturated fat or 5% daily value (DV) for saturated fat. What is the daily valu

e for saturated fat?
5g
20g
1g
10g
Mathematics
1 answer:
Leona [35]3 years ago
6 0

Answer:

The daily value for saturated fat is 20g.

Step-by-step explanation:

Percentage problems can be solved by rule of three

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too. In this case, the rule of three is a cross multiplication.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease. In this case, the rule of three is a line multiplication.

A percentage problem is an example where the relationship between the measures is direct.

The problem states that 1g is 5% of the daily value for saturated fat. The daily value(100%) for saturated fat is x, so:

1g - 5%

xg - 100%

5x = 100

x = \frac{100}{5}

x = 20g

The daily value for saturated fat is 20g.

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The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags.
Ivan

Answer and Step-by-step explanation:

First of all lets define  

X to be that the bags do contain large liquid amounts

+ to be that the test is not negative that is positive

From the question,  

P(x) = 3/100 = 0.03

P(+|X) = 0.91

P(+|X′) = 0.05

Probability of bag having a positive test =p(+)

= P(+|X) (P(x)) + P(+|X′)(P(X′))

= P(X′)) = 1 – 0.03 = 0.97

Inserting these values into these formulas  

0.91)(0,03) + (0.05)(0.97)

= 0.0273 + 0.0485

= 0.0758

The probability of the randomly selected bag having large liqid amount

=  P(X|+) = (0.91*0.03)/(0.91*0.03)(0.05*0.97)

= 0.0273/0.0273+0.0485

= 0.0273/0.0758

= 0.3602

the probability that this bag does not have large liquid amount

= p(X'|+) = 1 - P(X|+)

= 1 - 0.3602

= 0.6398

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Step-by-step explanation:

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