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jek_recluse [69]
3 years ago
5

He has 9 cans of white paint and 20 cans of blue paint.

Mathematics
1 answer:
Liula [17]3 years ago
3 0

Answer:

The answer to the problem is 1/2

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Step-by-step explanation:822.73

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Please help me out :)
Tomtit [17]
(f+g)(x)=5x-6+x^2-4x-8
A.(f+g)(x)=x^2+x-14 is correct:)
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What is 4 3/4 - 1/3 = in fraction form
pentagon [3]

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53/12

Step-by-step explanation:

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 The denominator of a fraction in simplest form is greater than the numerator by 3. If 7 is added to the numerator, and 5 added
jok3333 [9.3K]
So.. if we take the numerator to be say "a", then the denominator will be "a+3"

\bf \cfrac{a}{a+3}\textit{ if we add }\frac{1}{2}\textit{ we get then }\cfrac{a+7}{a+3+5}
\\\\\\
thus\implies \cfrac{a}{a+3}+\cfrac{1}{2}=\cfrac{a+7}{a+8}\\\\
-----------------------------\\\\
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3a^2+24a+3a+24=2a^2+14a+6a+42
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a^2+7a-18=0\implies (a+9)(a-2)=0
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a=
\begin{cases}
2\implies &\frac{2}{5}\\\\
-9\implies &\frac{3}{2}
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either of those two fractions will do
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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