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dimulka [17.4K]
2 years ago
11

I need help with this ASAP please

Mathematics
1 answer:
Doss [256]2 years ago
4 0

Answer:

c between 30 and 40 mins

Step-by-step explanation:

if you look at the graph that's the only line going down.

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Business people have determined that three fourths of the items on a mailing list will change in one year. A business has a mail
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1695 will be incorrect.
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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.

4 0
3 years ago
There was an activity planned outdoors, but it started to rain. They took a vote and 45 people of the 60 people wanted to stay i
Elza [17]

Answer:

15/60×100=25%

Step-by-step explanation:

the number of students wanted to go outside/the whole number of students×100

15/60×100=25%

4 0
3 years ago
Please help! solve.​
IRINA_888 [86]

Answer:

x=11

Step-by-step explanation:

6 0
3 years ago
How much simple interest is earned on $6000 principal, with an annual interest rate of 3.7% over 5 years?
diamong [38]
I = PRT
P (principal) = 6000
R (rate) = 3.7%.....u need decimal form = 0.037
T (time) = 5

I = (6000)(0.037)(5)
I = 1110.00 <===

u dont need years as a decimal...u need rate as a decimal
8 0
3 years ago
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