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Vera_Pavlovna [14]
3 years ago
11

True or false : if an individual or business breaks a contract, it automatically goes to court.

Mathematics
1 answer:
Alexus [3.1K]3 years ago
4 0

Answer:

It's false, only if the situation gets worse.

Step-by-step explanation:

On the off chance that a debate over a contract emerges and casual endeavors at determination fall flat, the foremost common next step may be a claim. On the off chance that the sum at issue is underneath a certain dollar figure (ordinarily $3,000 to $7,500 depending on the state), the parties may be able to resolve the issue in small claims court.

Please correct any mistakes in my answer :) Glad to help ya!

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If 5 Superscript 3 b minus 1 Baseline = 5 Superscript b minus 3, what is the value of b?.
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It sounds like you're given

5^{3b-1} = 5^{b-3}

Since the bases on either side match (5), the exponents must be equal. So

3b - 1 = b - 3

Solve for b :

3b - b = -3 + 1

2b = -2

b = -1

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Anya recorded the height in inches of 10 students in her class. The results are shown below.
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Step-by-step explanation:

I honestly have no clue what the answer is.

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3 years ago
The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviat
aleksley [76]

Answer:

4.5 hour and 2 hour

Step-by-step explanation:

Given: The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviation of two hours.

To Find: If each time was increased by one hour, what would be the new mean and standard deviation.

Solution:

let the total numbers entries of hours in survey be   =   \text{N}

each entry in survey be  =  \text{x}_{i}

mean of survey is     =     \mu  =3.5 \text{hours}

standard deviation is  =   \sigma = 2 \text{hours}

if each entry in survey is increased by one hour then,

each new entry in survey  is =  \text{x}_{i}+1

the new mean is\mu_{new}  = \frac{\text{sum of all hours}}{total number of entries}

                                                \frac{\text{x}_{1} +1+\text{x}_{2}+1 +....+\text{x}_{N}+1 }{N}

                                                \frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})+\text{N}\times1 }{N}

                                                \frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})}{N}+\frac{\text{N}\times1 }{N}

                                             \mu_{new} = \mu + 1=4.5 \text{hours}

now,

standard deviation is      \sigma_{new}= \sqrt{\sum_{1}^{N}\frac{(\text{x}_{inew}-\mu_{new})^{2}}{N}}

                                   \text{x}_{inew}= \text{x}_{i}+1

                                   \mu_{new}=\mu+1        

putting values,

                                                \sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}+1-\mu-1)^{2}}{N}}

                                                \sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}-\mu)^{2}}{N}}= \sigma

                                 \sigma_{new}        =2

new mean and standard deviation are 4.5 and 2 \text{hours}

                                               

5 0
3 years ago
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