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KonstantinChe [14]
3 years ago
9

A cable hangs between two poles of equal height and 34 feet apart. At a point on the ground directly under the cable and x feet

from the point on the ground halfway between the poles the height of the cable in feet is h(x)=10+(0.4)(x^1.5).
The cable weighs 18.2 pounds per linear foot.
Find the weight of the cable.
Mathematics
1 answer:
Sergio039 [100]3 years ago
8 0

Answer:

5000

Step-by-step explanation:

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3/6 possible out comes for heads and tails
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acute

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obtuse

Step-by-step explanation:

those are the answers

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What is 35+20/-4 times 7=?
saw5 [17]
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There is a mound of g pounds of gravel in a quarry. Throughout the day, 300 pounds of gravel are added to the mound. Two orders
galina1969 [7]

Answer:

g+300-700-700=1500, g=2600

Step-by-step explanation:

At the start of the day there are g pounds of gravel in the quarry.

Then it says that there were 300 pounds added to the mound, so at this point the pounds of gravel in the mound are now g+300.

Then two orders of 700 pounds are sold and removed from the mound. That means they removed 700 pounds from the mound 2 times. Therefore at this point the pounds of gravel in the mound are now g+300-700-700.

Then it says that at this point, the amount of pounds in the mound are 1500.  Hence we get

g+300-700-700=1500

To solve for g we just add 700 two times to both sides of the equation, and subtract 300 from both sides of the equation, getting:

g+300-700-700+700+700-300=1500+700+700-300

And so g=2600

4 0
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.

4 0
3 years ago
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