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damaskus [11]
2 years ago
12

Convert 3.12 mm into cm, m and km.​

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
7 0

Answer:

3.12mm= 31,2cm

3.12= 0.312m

3.12 = 0.000312km

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Y= -3/5 - 1

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if you start on the y intercept an go down once (-1)

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positive 5 so go right 5 times it lands on the point :)

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Which of the following angle relationships in △ABC are correct? Select all that apply. There is a triangle ABC in which the leng
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  A, B, D

Step-by-step explanation:

The sides in order from smallest to largest are ...

  AC, BC, AB

Then the opposite angles in order from smallest to largest are ...

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Let's consider the choices:

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2 years ago
Dragon drop each expression two correctly classify it as having a positive or negative product
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Step-by-step explanation:

We multiply it to find the product

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negative times negative = positive

(-\frac{1}{3})(-\frac{1}{3})= \frac{1}{9}

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3 years ago
An article reports the following data on yield (y), mean temperature over the period between date of coming into hops and date o
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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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3 years ago
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Step-by-step explanation:

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