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Margarita [4]
3 years ago
9

Given f(x) =1 - 3/4x find the following.

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
5 0

Answer:

a. f(-12) = 10

b. f(0) = 1

c. f(4) = -2

Step-by-step explanation:

In order to find the function of each of these numbers, you just need to replace them in the function as x and then solve:

<u>a.</u> f(-12) = 1 - 3/4(-12)

 - f(-12) = 1 - (-36/4)

 - f(-12) = 1 + 9

 - f(-12) = 10

<u>b.</u> f(0) = 1 - 3/4(0)

 - f(0) = 1 - 0

 - f(0) = 1

<u>c.</u> f(4) = 1 - 3/4(4)

 - f(4) = 1 - 12/4

 - f(4) = 1 - 3

 - f4) = -2

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Answer:

Null hypothesis is rejected. Standard deviation is > 8 miles per hour

Step-by-step explanation:

σ [Population standard deviation] = 8 , s [sample standard deviation] = s

n [no of items] = 8

H0 [Null] : σ = 8 ; H1 [Alternate - Right Tail] : σ > 8

χ2 = (n - 1) . s^2 / σ^2

= 49 x (10.5)^2 / 82 = 5402.25 / 64

χ2 = 84.410

df [degree of freedom] = n -1 = 50 - 1 = 49

P value (χ^2 49 > 84.410) = 0.00125

p = 0.0013

p < α ie 0.05

So, H0 is rejected

Hence we state that standard deviation is > 8 miles per hpur

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3 years ago
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Consider the case0502 data from Sleuth3. &lt;&lt;&lt; This is the data. Sleuth3 is preloaded into R studio.
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Answer:

Consider the following calculations

Step-by-step explanation:

The complete R snippet is as follows

install.packages("Sleuth3")

library("Sleuth3")

attach(case0502)

data(case0502)

## plot

# plots

boxplot(Percent~ Judge, data=case0502,ylab="Values",

main="Boxplots of the Data",col=c(2:7,8),horizontal=TRUE)

# perform anova analysis

a<- aov(lm(Percent~ Judge,data=case0502))

#summarise the results

summary(a)

### we can use the independent sample t test here

sp<-case0502[which(case0502$Judge=="Spock's"),]

nsp<-case0502[which(case0502$Judge!="Spock's"),]

## perform the test    

t.test(sp$Percent,nsp$Percent)

The results are CHECK THE IMAGE ATTACHED

b)

> summary(a)

Df Sum Sq Mean Sq F value Pr(>F)

Judge 6 1927 321.2 6.718 6.1e-05 *** as the p value is less than 0.05 , hence there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge

Residuals 39 1864 47.8

---

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

c)

t.test(sp$Percent,nsp$Percent)

  Welch Two Sample t-test

data: sp$Percent and nsp$Percent

t = -7.1597, df = 17.608, p-value = 1.303e-06 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-19.23999 -10.49935

sample estimates:

mean of x mean of y

14.62222 29.49189

6 0
3 years ago
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