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IRINA_888 [86]
3 years ago
12

Heidi has $4.55. She needs to buy bolts at the hardware store . Bolts cost$0.13 each. How many bolts can Heidi buy.

Mathematics
2 answers:
pogonyaev3 years ago
7 0
Hey there!

One bolt = $0.13
Hence, divide 4.55 by 0.13

4.55 \div 0.13 = 35

Answer : Heidi can buy 35 bolts.

Hope this helps. - M
Rom4ik [11]3 years ago
3 0
1 bolt = $0.13

Number of bolts she can buy = $4.55 ÷ $0.13 = 35

Answer: 35
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d soon is the answer

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What is the answer ?
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A statistician chooses 27 randomly selected dates, and when examining the occupancy records of a particular motel for those date
Anni [7]

Answer:

Confidence interval variance [21.297 ; 64.493]

Confidence interval standard deviation;

4.615, 8.031

Step-by-step explanation:

Given :

Variance, s² = 34.34

Standard deviation, s = 5.86

Sample size, n = 27

Degree of freedom, df = 27 - 1 = 26

Using the relation for the confidence interval :

[s²(n - 1) / X²α/2, n-1] ; [s²(n - 1) / X²1-α/2, n-1]

From the chi distribition table :

X²α/2, n-1 = 41.923 ; X²1-α/2, n-1 = 13.844

Hence,

[34.34*26 / 41.923] ; [34.34*26 / 13.844]

[21.297 ; 64.493]

The 95% confidence interval for the population variance is :

21.297 < σ² < 64.493

Standard deviation is the square root of variance, hence,

The 95% confidence interval for the population standard deviation is :

4.615 < σ < 8.031

3 0
3 years ago
William and his puppy Rover are out for a morning jog at a comfortable pace of 10 minutes per mile. If they keep up this pace fo
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Answer:

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8 0
3 years ago
What is the square root of 95.54
vladimir2022 [97]
Well this is simple a calculator type problem...but if you are curious as the the algorithm used by simple calculators and such...

They use a Newtonian approximation until it surpasses the precision level of the calculator or computer program..

A newtonian approximation is an interative process that gets closer and closer to the actual answer to any mathematical problem...it is of the form:

x-(f(x)/(df/dx))

In a square root problem you wish to know:

x=√n  where x is the root and n is the number

x^2=n

x^2-n=0

So f(x)=x^2-n and df/dx=2x so using the definition of the newton approximation you have:

x-((x^2-n)/(2x)) which simplifies further to:

(2x^2-x^2+n)/(2x)

(x^2+n)/(2x), where you can choose any starting value of x that you desire (though convergence to an exact (if possible) solution will be swifter the closer xi is to the actual value x)

In this case the number, n=95.54, so a decent starting value for x would be 10.

Using this initial x in (x^2+95.54)/(2x) will result in the following iterative sequence of x.

10, 9.777, 9.774457, 9.7744565, 9.7744565066299210578124802523397

The calculator result for my calc is: 9.7744565066299210578124802523381

So you see how accurate the newton method is in just a few iterations. :P


5 0
3 years ago
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