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Iteru [2.4K]
3 years ago
12

A local grocery store decides to offer a free piece of fresh fruit (banana or apple) to all shoppers in the produce department.

The store is conducting an observational study to determine which type of fruit is selected more often. At the end of the first day, the store found that twice as many shoppers select an apple.
The grocery store then repeats the observational study for 14 days. All studies yield similar results. What generalization can be made from the results of this study?

A.
Given the choice of a banana or an apple, twice as many shoppers will select an apple.
B.
The results are inconclusive; therefore, a generalization cannot be made regarding which type of fruit is preferred by most shoppers.
C.
There is not enough information to generalize the study’s results.
D.
Given the choice of any type of fruit, twice as many shoppers will select an apple.
Mathematics
2 answers:
Alisiya [41]3 years ago
7 0

Step-by-step explanation:

A. Given the choice of a banana or an apple, twice as many shoppers will select an apple.

hope it helps!

almond37 [142]3 years ago
3 0

Answer:

A.

Step-by-step explanation:

If the results are similar (A) should be your answer!

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I need c and d answered badly
Kisachek [45]

C. translated down 2 units

---The "-2" located on the outside of f(x) tells us that the y-values are being changed. With that, the graph can be moved up or down. The presence of the negative/subtraction sign tells us that the graph is moved down.

D. translated right 4 units

---The "-4" located with the x in f(x) tells us that the x-values are being changed. The only tricky thing about this is that the direction of the movement is actually the opposite of the sign. So with the negative/subtraction sign, the graph is moved to the right instead of the left.

Hope this helps!

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2 years ago
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If RON becomes 181514 what becomes 122122?
goldfiish [28.3K]

Answer:

B. LUV

Step-by-step explanation:

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7 0
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An article reports the following data on yield (y), mean temperature over the period between date of coming into hops and date o
skelet666 [1.2K]

Answer:

x1=c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)

x2=c(30,42,47,47,43,41,48,44,43,50,56,60)

y=c(210,110,103,103,91,76,73,70,68,53,45,31)

mod=lm(y~x1+x2)

summary(mod)

R output: Call:

lm(formula = y ~ x1 + x2)

Residuals:  

   Min      1Q Median      3Q     Max

-41.730 -12.174   0.791 12.374 40.093

Coefficients:

        Estimate Std. Error t value Pr(>|t|)    

(Intercept) 415.113     82.517   5.031 0.000709 ***  

x1            -6.593      4.859 -1.357 0.207913    

x2            -4.504      1.071 -4.204 0.002292 **  

---  

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  

Residual standard error: 24.45 on 9 degrees of freedom  

Multiple R-squared: 0.768,     Adjusted R-squared: 0.7164  

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a).  y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

c).  y =415.113 +(-6.593)*21.3 +(-4.504)*43 =81.0101

residual =68-81.0101 = -13.0101

d). F=14.9

P=0.0014

There is convincing evidence at least one of the explanatory variables is significant predictor of the response.

e).  newdata=data.frame(x1=21.3, x2=43)

# confidence interval

predict(mod, newdata, interval="confidence")

#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

      fit      lwr      upr

1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f).  #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

      fit      lwr      upr

1 81.03364 14.19586 147.8714

95% PI=(14.20, 147.87)

g).  No, there is not evidence this factor is significant. It should be dropped from the model.

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