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shepuryov [24]
3 years ago
6

Can someone give me advice please..I'm having a panic attack because a car hit my car when i was turning on a yellow arrow and i

t was clear but this brown jeep speeded to hit my car. how do i fix it without my dad knowing. i am willing to pay a lot to fix it but i just don't want to get in trouble and I don't want him finding out :(​
Mathematics
2 answers:
vesna_86 [32]3 years ago
8 0
Find a friend with a garage and hide it there if you don’t have any with a garage then just hide it somewhere he’d never look or has never looked but if he finds it just explain what happened and he might understand

If this done help I’m sorry! God rest your soul
rosijanka [135]3 years ago
4 0

Answer:

keep the car at a friends house untill you find a place to get it fixed at... or keep it at a trustworthy family members house.

If these don't help then im sorry.

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What is the solution of the equation 4.5+m=10
drek231 [11]

Answer:

The first thing that we need to do is subtract 4.5 from both sides because that would removed the 4.5 from the left side and leave us with m by itself.  So we get the following m=5.5 and that is our solution.

<u><em>Hope this helps!  Let me know if you have any questions</em></u>

6 0
1 year ago
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The Pythagoreans discovered irrationals in about the ____th Century BC.
Naddika [18.5K]

Answer:

5th

Step-by-step explanation:

3 0
3 years ago
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A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sam
soldi70 [24.7K]

Answer:

a) 52.8-2.02\frac{3.9}{\sqrt{40}}=51.554    

b) 52.8+2.02\frac{3.9}{\sqrt{40}}=54.046    

c) 52.8-2.71\frac{3.9}{\sqrt{40}}=51.129  

d) 52.8+2.71\frac{3.9}{\sqrt{40}}=54.471  

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

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 =52.8 represent the sample mean for the sample

\mu population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size  

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

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 mean and the sample deviation we can use the following formulas:  

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

df=n-1=40-1=39

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 table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that t_{\alpha/2}=2.02

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

52.8-2.02\frac{3.9}{\sqrt{40}}=51.554    

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

52.8+2.02\frac{3.9}{\sqrt{40}}=54.046    

So on this case the 95% confidence interval would be given by (51.554;54.046)    

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that t_{\alpha/2}=2.71

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

52.8-2.71\frac{3.9}{\sqrt{40}}=51.129  

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

52.8+2.71\frac{3.9}{\sqrt{40}}=54.471  

So on this case the 99% confidence interval would be given by (51.129;54.471)    

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

3 0
3 years ago
Basic math
LenKa [72]

Answer:

The quotient contains a terminating decimal and The quotient is a whole number less than 11.

Step-by-step explanation:

To answer this one, it's mandatory to remember that quotient, is the outcome of a ratio: a number (r) over another (s) (different than 0). In this case:\frac{81}{918}. So q is equal to =0.08823529411.

Analyzing the number: 0.08823529411

This is not a repeating decimal, but it is a terminating decimal for it has an end.

The quotient is also a whole number less than 11.

The Whole Set of numbers is made up of the following numbers W ={0,1,2,...} and 0 < 11. Therefore it is true.

8 0
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If y varies directly as (x+3) and inversely as (x-3) and y=21 when x=4, what is the equation for y in terms of x?
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\bf \qquad \qquad \textit{double proportional variation}&#10;\\\\&#10;\begin{array}{llll}&#10;\textit{\underline{y} varies directly with \underline{x}}\\&#10;\textit{and inversely with \underline{z}}&#10;\end{array}\implies y=\cfrac{kx}{z}\impliedby &#10;\begin{array}{llll}&#10;k=constant\ of\\&#10;\qquad  variation&#10;\end{array}\\\\&#10;-------------------------------\\\\

\bf \textit{\underline{y} varies directly as (x+3) and inversely as (x-3)}\implies y=\cfrac{k(x+3)}{x-3}&#10;\\\\\\&#10;\textit{we also know that }&#10;\begin{cases}&#10;y=21\\&#10;x=4&#10;\end{cases}\implies 21=\cfrac{k(4+3)}{4-3}\implies 21=\cfrac{k(7)}{1}&#10;\\\\\\&#10;\cfrac{21}{7}=k\implies 3=k\qquad thus\qquad \qquad \boxed{y=\cfrac{3(x+3)}{x-3}}
6 0
3 years ago
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