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hjlf
3 years ago
6

Write and simplify the equation you would use to solve the problem below.

Mathematics
1 answer:
Taya2010 [7]3 years ago
7 0
12s + 5 = 77
12s = 77 - 5 = 72
s = 72/12
s = 6 ...//

Ans = C ... 12x + 5 = 77 ...
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Jared measured a house and its lot and made a scale drawing. The scale of the drawing was 5 millimeters : 3 meters. The actual l
alukav5142 [94]

Answer:

115mm

Step-by-step explanation:

5/x = 3/69

5/x = 1/23

5*23 = x

115 = x

4 0
3 years ago
3. Define a variable, write an equation, and solve the problem. Carla began a
faust18 [17]

Answer:

Carla has been running 7 miles each day for 13 days.

Step-by-step explanation:

Carla ran 3 miles on her first day, 5 miles on her second day and then 7 miles each day onward.

Let the number of days she ran 7 miles each day = x

Total distance run by Carla = 3 + 5 + 7(x)

                                             = 8 + 7x

If her log book shows that she has run total distance = 99 miles

Equation representing her total run will be,

8 + 7x = 99

7x = 99 - 8

7x = 91

x = \frac{91}{7}

x = 13 days

Therefore, Carla has been running 7 miles each day for 13 days.

3 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.

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3 years ago
What is the Slope?<br> 4x + y = 1
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Answer:

m=-4

Step-by-step explanation:

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2 years ago
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$456 is divided between the telephone and electricity bills in a ratio of 2 : 1. What amount goes to the telephone bill?
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$304 is the answer because 456/3 and then multiply by 2
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