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Vanyuwa [196]
2 years ago
11

Using the given points determine ∆y. (-3,-5) and (0,10)

Mathematics
1 answer:
ryzh [129]2 years ago
4 0
If the question is only asking for the difference between the y in the two points, then the difference can be simply calculated as y2 - y1 as follows:
<span>∆y = y2 - y2 = 10-(-5) = 15 units

If the two given points are to be used to determine the slope of a line, then the slope can be calculated as follows:
slope = (y2-y1) / (x2-x1) = (10--5) / (0--3) = 15/3 = 5</span>
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6. A prism has bases that are equilateral triangles with sides lengths of 10 inches and a length of 30 inches.
vova2212 [387]

Answer:

The volume of the prism is 1,299\ in^{3}

Step-by-step explanation:

we know that

The volume of the prism is equal to

V=BL

where

B is the area of the triangular base

L is the length of the prism

we have

L=30\ in

<em>Find the area of the base B</em>

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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
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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