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kogti [31]
3 years ago
15

How to round 106.327 to nearest hundredths explain

Mathematics
2 answers:
scoray [572]3 years ago
8 0

Answer:

106.33

Step-by-step explanation:

after decimal you have .3 tenths 2 hundredths 7 thousandths

to round to the nearest hundredths look at thousandths 7 becasuse is greater then 5 then you round up

106.327 is rounded to 106.33

butalik [34]3 years ago
4 0

Answer:

your answer is 106.3

Step-by-step explanation: so first you do this

106.327\\3 =tenths\\2= hundredths

so when you round  if it is 5 or above you round up if it is 4 or below leave it alone

so the final answer is 106.3

when you round the 27 is removed and this is what you have left.

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A cone with a diameter
kkurt [141]

Answer:

I think it 3.15 if it wrong I'm not even here

3 0
3 years ago
A classroom is 26ft wide, 32ft long, and 9ftt high. what is the volume of the room in cubic feet?
julsineya [31]

Answer:

7488 ft^3

Step-by-step explanation:

26x32x9 = 7488 cubic ft

3 0
3 years ago
Read 2 more answers
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb
ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

6 0
3 years ago
Help geometry what I m jlf <br> Picture provided
Zolol [24]

Answer:

\angle\,JLF = 114^o  which agrees with option"B" of the possible answers listed

Step-by-step explanation:

Notice that in order to solve this problem  (find angle JLF) , we need to find the value of the angle defined by JLG and subtract it from 180^o, since they are supplementary angles. So we focus on such, and start by drawing the radii that connects the center of the circle (point "O") to points G and H, in order to observe the central angles that are given to us as 90^o and 138^o. (see attached image)

We put our efforts into solving the right angle triangle denoted with green borders.

Notice as well, that the triangle JOH that is formed with the two radii and the segment that joins point J to point G, is an isosceles triangle, and therefore the two angles opposite to these equal radius sides, must be equal. We see that angle JOH can be calculated by : 360^o-90^o-138^o=132^o

Therefore, the two equal acute angles in the triangle JOH should add to:

180^o-132^o=48^o resulting then in each small acute angle of measure 24^o.

Now referring to the green sided right angle triangle we can find find angle JLG, using: 180^o-24^o-90^o=66^o

Finally, the requested measure of angle JLF is obtained via: 180^o-66^o=114^o

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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