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Savatey [412]
3 years ago
6

Need the answer for numbers 4,8,and 12

Mathematics
1 answer:
anzhelika [568]3 years ago
8 0

9514 1404 393

Answer:

  4) -438.75

  8) 539.68

  12) 22.54

Step-by-step explanation:

Put the values in place of the corresponding variables and do the arithmetic.

__

4) 6xy = 6(-7 1/2)(9 3/4) = -6(7.5)(9.75) = -438.75

__

8) I = prt = $2442 × 0.085 × 2.6 = $539.68

__

12) I = prt = ($578)(0.0325)(1.2) = $22.54

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Determine how many lines of symmetry each object has. Then determine weather each object has rotational symmetry
Dafna1 [17]

Answer:

2: yes

Step-by-step explanation:

5 0
3 years ago
The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds
Bumek [7]
Answer: 11.5% 

Explanation:


Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds. 

Now, we get the z-score for 720 seconds by the following formula:

\text{z-score} =  \frac{x - \mu}{\sigma}

where 

t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}

So, the z-score of 720 seconds is given by:

\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

P(t \ \textgreater \  720) 
\\ = P(z \ \textgreater \  1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\  \boxed{P(t \ \textgreater \  720)  = 0.115}

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice. 
8 0
3 years ago
A movie theater sold 1,246 tickets this weekend. If 314 tickets were sold on Friday and 529 tickets were sold on sunday
Lisa [10]

Answer:

Saturday=403 tickets

Step-by-step explanation:

1246-314-529=403

6 0
3 years ago
The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor
nadya68 [22]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We can use as an estimator for p \hat p =0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

3 0
3 years ago
in a game a team has 7 wins to 5 losses. of the team doesn't lose anymore how many games must the team win to have and win :lose
nadezda [96]
I don’t know but hope you have a good day
3 0
2 years ago
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