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lara [203]
3 years ago
9

A reasonable estimate for the mass of a baby would be 7 __ pounds

Mathematics
2 answers:
zaharov [31]3 years ago
7 0
Maybe 7.4 pounds idk sorry if I’m wrong
Ahat [919]3 years ago
6 0
Well if we’re looking at average here... a good estimate for a baby at birth is 7.5 pounds.
So in the blank put .5 because 6 is the pounds and .5 is the ounces.
Hope this helps!
~Brooke❤️
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Is 3/7 bigger than 1/4
dmitriy555 [2]
Yes! it is bigger than 
8 0
2 years ago
Here are summary statistics for randomly selected weights of newborn​ girls: nequals=174174​, x overbarxequals=30.930.9 ​hg, seq
MaRussiya [10]

Answer:

The 95% confidence interval would be given by (29.780;32.020)  

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=30.9 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=174-1=173

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that t_{\alpha/2}=1.97, this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.  

Now we have everything in order to replace into formula (1):

30.9-1.97\frac{7.5}{\sqrt{174}}=29.780    

30.9+1.97\frac{7.5}{\sqrt{174}}=32.020

So on this case the 95% confidence interval would be given by (29.780;32.020)    

The value 29.6 is not contained on the interval calculated.

5 0
3 years ago
160.1/2 ÷ 70.1/3<br>I Mark brainlies ​
Mama L [17]

Answer:

3.42582025

Step-by-step explanation:

According to a high-standard calculator, this is the answer.

Hope I helped:)!!

6 0
2 years ago
Find the range 7,6,5,9,5,8,2 please include work!!
ArbitrLikvidat [17]

Answer:

7.

Step-by-step explanation:

Range = highest value - lowest

= 9 - 2

= 7.

5 0
3 years ago
F(t) = t-1<br> g(t)= t^3 + 4t<br> Find (f -g)(t)
Annette [7]

Answer:

Step-by-step explanation:

t - 1 - t^3 - 4t

-t^3 - 3t

5 0
3 years ago
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