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zalisa [80]
3 years ago
7

Peanut allergies are becoming increasingly common in Western countries. Some evidence points to the timing of peanuts! first int

roduction in the diet as an influential factor, raising the question of whether pediatricians should recommend early exposure or avoidance. A study enrolled infants with a diagnosed peanut allergy and randomly assigned them to either completely avoid peanuts or consume peanuts in small amounts regularly until they reached 60 months of age. At the end of the study, 18 of the 51 infants who had avoided peanuts were still allergic to peanuts. In contrast, 5 of the 47 infants who had consumed peanuts were still allergic to peanuts. Do the data indicate that one approach is more beneficial? Follow the four‑step process.
STATE: What is the question we are asking?
A. Does early exposure to peanuts better reduce peanut allergy in children with a known allergy to peanuts?
B. Does early exposure to peanuts in small amounts or complete avoidance of peanuts better reduce peanut allergy in children with a known allergy to peanuts?
C. Should all children under the age of 60 months avoid peanuts or only eat them in small amounts?
D. Should all children under 60 months eat peanuts regularly in small amounts?
Mathematics
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

<u>B. Does early exposure to peanuts in small amounts or complete avoidance of peanuts better reduce peanut allergy in children with a known allergy to peanuts?</u>

<u>Step-by-step explanation:</u>

Remember we are told there's already a question <em>whether pediatricians should recommend</em> <em>early exposure or avoidance.</em> Thus, the research question should answer these two i<em>ssues</em> (<em>early exposure or avoidance)</em>.

Hence, the question we are asking is: Does early exposure to peanuts in small amounts or complete avoidance of peanuts better reduce peanut allergy in children with a known allergy to peanuts?

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Several years ago, Matt built a rectangular flower bed at his house. The width of the flower bed was 14 inches less than the len
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D. 24.47 inches
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3 years ago
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Roisin is a price analyst for a food distributor. One week she put together bids for 680 cases, 507 cases, 830 cases, 391 cases,
sineoko [7]

Answer: The average number of cases per bid is <u>599</u>.

Step-by-step explanation:

Given: One week she put together bids for 680 cases, 507 cases, 830 cases, 391 cases, and 587 cases.

Total cases = 680 +  507 + 830 + 391+587

= 2995

Number of bids = 5

Now, average number of cases per bid = \dfrac{\text{Total cases }}{\text{Number of bids}}

=\dfrac{2995}{5}\\\\=599

Hence, the average number of cases per bid is <u>599</u>.

6 0
3 years ago
The sum of three consecutive even numbers is one hundred sixty-two. What is the smallest of the three numbers?
zepelin [54]
Since the numbers are consecutive, they must be near each other in size, and therefore near one-third of 168. And since the lower number is -2 off the middle, and the upper number is +2 from the middle, the middle number must be exactly one-third of 168—otherwise the sum of the three could never be 168. Labelling the three consecutive numbers as x, y, and z:

x + y + z = 168

(y-2) + y + (y+2) = 168.

3y = 168

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therefore the smallest number (x) = 54.
436 viewsView upvotes

1




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3 0
3 years ago
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A true-false quiz with 10 questions was given to a statistics class. Following is the probability distribution for the score of
disa [49]

Answer:

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

X         5       6         7       8        9        10

P(X)   0.05   0.15  0.33  0.28   0.12   0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:  

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :  

E(X^2)=\sum_{i=1}^n X^2_i P(X_i)  

And using the formula we got:  

E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46  

Then we can find the variance with the following formula:  

Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096  

And then the standard deviation would be given by:  

Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229  

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andrew-mc [135]

Answer:

B, 11000

Step-by-step explanation:

assuming a 365 day year...

10500*(1+.15)^{\frac{123}{365}}= 11006.35813

which I guess is close enough to 11000

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