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1 year ago

13

A goal is to get at least 10% of daily calories from polyunsaturated fats as are found in various nuts, oils, and fish. One serv

ing of walnuts (about 15 pieces) has about 20 g of polyunsaturated fat. Each fat gram has 9 calories. About what percent of daily calories does one serving of walnuts provide, assuming a 2,000-calorie diet?

1 answer:

natka813 [3]1 year ago

6 0

To get 10% of your calories from polyunsaturated fats you would need to eat 200 calories worth. So 1 serving of walnuts would give you 180 calories. So one serving of walnuts would be 9% of your daily calories.

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During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

melomori [17]

Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ $b+c=20$
Equation [2] ⇒ $1.50b+5c=54.50$

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; $b$ and $c$ , that we need to solve, then this becomes a simultaneous equation problem.

First, rearrange Equation [1] to make either $b$ or $c$ the subject:
$b+c=20$
$b=20-c$

Then we substitute $b=20-c$ into Equation [2]
$1.50b+5c=54.50$
$1.50(20-c)+5c=54.50$
$30-1.50c+5c=54.50$
$5c-1.5c=54.50-30$
$3.5c=24.50$
$c=7$

Now we know the value of $c$ which is $c=7$ , substitute this value into $b=20-c$ we have $b=20-7=13$

Answer:
Numbers of books = 13
Numbers of CDs = 7

5 0

2 years ago

Solve the inequality 47.75 + x Less-than-or-equal-to 50 to determine how much more weight can be added to Li's suitcase without

myrzilka [38]

Answer:

x is less-than-or-equal-to 2.25 (x ≤ 2.25)

Step-by-step explanation:

We can write down the inequality that represents the weight Li can add without going over the 50 pound limit:

47.75 + x ≤ 50

If we solve for x we have:

47.75 + x ≤ 50

x ≤ 50 - 47.75

x ≤ 2.25

Therefore, the weight Li can add to the suitcase is less-than-or-equal-to 2.25

6 0

1 year ago

Read 2 more answers

Mr. Dennis travels 27.2 miles to work each day.

galina1969 [7]

Answer:

g5v869v96r9v

rc55vdr66ves5c5sxStep-by-step explanation:

6 0

1 year ago

A radio station receives 3815 phone calls in 35 min. How many calls does it get in 1 hr? Show Your Work.

Hatshy [7]

In 1 min it receives 3815/35=109 so in 1 hour it receives 109x60=6540 calls

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1 year ago