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aliya0001 [1]
2 years ago
5

One serving (1 cup) from the fruits group is equal to 1 cup of fruit, 1 cup of 100% fruit juice, or 1/2 cup of dried fruit. Why

is the serving size for dried fruit smaller than the serving size for other forms of fruit?
Mathematics
1 answer:
Dmitrij [34]2 years ago
6 0

Answer:

Because the dried fruit contains less water

Step-by-step explanation:

Drying is a phenomenon whereby heat is used to remove moisture from a material.

Fruit juice contains about 85% water, that means the remaining fruit content is 25%

The average water content of fruits is about 89%, this means that the remaining fruit content is 11%.

Dried fruit contains about 20% water which means that the remaining fruit content is 80%.

If we assume that each cup serving weighs 1kg this implies that;

The real fruit content (vitamins, minerals etc) masses are;

Fruit juice = 0.25kg

Fruit = 0.11kg

Dried fruit = 0.8kg

We can see that the mass of the fruit content in the dried fruit is twice more than than of the fruit and fruit juice individually. This is why half cup of dry fruit is equated to one cup of the other two.

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Capucine played a game where she earned 179 points from building 11 museums and 4 libraries. She earns 4 more points per museum
Korolek [52]

Capucine played a game where she earned 179 points from building 11 museums and 4 libraries.

Lets assume  x points she earn per library

She earns 4 more points  for museum than  library.

4 more points per museum per x .  so the points earned per museum is x+4

she earned 179 points from 11 museums and 4 libraries.

So the equation becomes

11 (x+4) + 4x = 179

11x +44 + 4x = 179

15x + 44 = 179 ( subtract 44 from both sides)

15x= 135 (divide by 15 from both sides)

x = 9

She earns 9 points per library

We know points per museum = x+4 = 9+4 = 13

She earns 13 points per museum .

5 0
2 years ago
Which statement correctly explains how Andre could find the solution to the following system of linear equations using eliminati
grin007 [14]

umm ,this is too long could you maybe putt it in smaller form so I can answer

7 0
1 year ago
Rich is driving from Philadelphia to Pittsburgh at 70 mph and Michelle is driving from Pittsburgh to Philadelphia at 65 mph. If
Margaret [11]
The distance from <span>Philadelphia to Pittsburgh is 305 miles.

Both are travelling towards each other. When they will meet the sum of distance traveled by both would be 305 miles, as together they have covered the entire distance between the two cities.

We know that:

Distance = Speed x Time

We know the distance. We know the speeds. Since they are travelling towards each other, the overall speed will be the sum of speeds of Rich and Michelle. So we can say with a speed of 135 mph they have to cover 305 miles.

305 = 135 x Time

Time = 305/135 = 2.25 hours = 2 hours and 15 minutes

So, they will meet after 2 hours and 15 minutes.

So, C option is the correct answer</span>
5 0
2 years ago
Read 2 more answers
A study investigated about 3000 meals ordered from Chipotle restaurants using the online site Grubhub. Researchers calculated th
Allisa [31]

Answer:

21.186%

Step-by-step explanation:

z = (x-μ)/σ,

where

x is the raw score = 2400mg

μ is the population mean = 2000mg

σ is the population standard deviation = 500mg

z = 2400 - 2000/500

z = 0.8

Probability value from Z-Table:

P(x<2400) = 0.78814

P(x>2400) = 1 - P(x<2400) = 0.21186

Converting to percentage:

0.21186 × 100

= 21.186%

Therefore, the percent of the meals

ordered that exceeded the recommended daily allowance of 2400 mg of sodium is 21.186%

7 0
2 years ago
Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and a
zaharov [31]

Answer:

From the hypothesis test carried out, the p-value obtained < the significant level at which the test was performed at, hence, the company can conclude that the New Club has increased average driving distance compared with the Leading Club.

Step-by-step explanation:

The results of drives of 10 players using the new and the leading club is presented thus

New club | Leading club

236.4 | 237.2

202.5 | 200.4

245.6 | 240.8

257.4 | 259.3

223.5 | 218.9

205.3 | 200.6

266.7 | 258.9

240.0 | 236.5

278.9 | 280.5

211.4 | 206.5

The test is to check if the New Club has increased average driving distance compared with the Leading Club.

Since the two sets of average drives belong to the same set of people, a t-test is appropriate.

To do this test, we first take the difference in these average drives for New Club and Leading Club.

New club | Leading club | Difference

236.4 | 237.2 | -0.8

202.5 | 200.4 | 2.1

245.6 | 240.8 | 4.8

257.4 | 259.3 | -1.9

223.5 | 218.9 | 4.6

205.3 | 200.6 | 4.7

266.7 | 258.9 | 7.8

240.0 | 236.5 | 3.5

278.9 | 280.5 | -1.6

211.4 | 206.5 | 4.9

The mean of differences = (sum of the differences) ÷ (sample size)

Sum of the differences = -0.8 + 2.1 + 4.8 + -1.9 + 4.6 + 4.7 + 7.8 + 3.5 + -1.6 + 4.9 = 28.1

Sample size = 10

mean = (28.1/10) = 2.81

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = 13.0321+0.5041+3.9601+22.1841+3.2041+3.5721+24.9001+0.4761+19.4481+4.3681

= 95.649

σ = √[Σ(x - xbar)²/N] = √(95.649/10) = 3.093

To do an hypothesis test, we need to first define the null and alternative hypothesis

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

The test is to check if the New Club has increased average driving distance compared with the Leading Club.

The null hypothesis would then be that there is no significant evidence that the New Club has increased average driving distance compared with the Leading Club.

That is, the average driving distance with the new club isn't higher than the average driving distance with the leading club or the new club's average drive is lesser or equal to the average drive with the leading club or the difference of the average driving distance with the new club and the average driving distance with the leading club is less than or equal to 0.

The alternative hypothesis in the other hand would be that there is significant evidence that the New Club has increased average driving distance compared with the Leading Club.

That is, the difference of the average driving distance with the new club and the average driving distance with the leading club is greater than 0.

If the population difference of the average driving distance with the new club and the average driving distance with the leading club is μ

Mathematically,

The null hypothesis is represented as

H₀: μ ≤ 0

The alternative hypothesis is given as

Hₐ: μ > 0

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = mean difference between two sets of data obtained from the sample = 2.81

μ₀ = 0

σₓ = standard error = (σ/√n)

where n = Sample size = 10

σₓ = (σ/√n) = (3.093/√10) = 0.978

t = (2.81 - 0) ÷ 0.978

t = 2.87

checking the tables for the p-value of this t-statistic

Degree of freedom = df = 10 - 1 = 10 - 1 = 9

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing only in one direction. (Greater than 0 direction)

p-value (for t = 2.87, at 0.05 significance level, df = 9, with a one tailed condition) = 0.009238

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.009238

0.009238 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the difference of the average driving distance with the new club and the average driving distance with the leading club is greater than 0.

Hope this Helps!!!

7 0
2 years ago
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