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Nimfa-mama [501]
3 years ago
9

If ethans monthly expenses are 1160 and his debt to income ratio is 0.8 what is his monthly salary

Mathematics
1 answer:
lukranit [14]3 years ago
7 0

Answer:

\(1450\)

Step-by-step explanation:

Based on the given conditions, formulate: \frac{1160}{0.8}

Convert decimal to fraction:\(\dfrac{1160} {\dfrac{8}{10}}\)

Divide a fraction by multiplying its reciprocal:\(1160 \times \dfrac{10}{8}\)

Cross out the common factor: \(145 \times 10\)

Calculate the product or quotient: \(1450\)

get the result: \(1450\)

Answer: \(1450\)

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An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

Step-by-step explanation:

Previous concepts

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
I not sure what it is? Thank you to who ever help me
Vadim26 [7]
I think b I might be wrong
8 0
3 years ago
Agatha is five times older than her son bob. three years ago she was nine times older. how old is agatha?
Naya [18.7K]
Write an equation system based on the problem
"Agatha is five times older than her son, Bob" could be written as:
  a = 5b.......(first equation)
"Three years ago, she was nine times older" could be written as:
  a - 3 = 9(b - 3)......(second equation)

Solve the equation system
To find the value of b, substitute 5b as a to the second equation
a - 3 = 9(b - 3)
5b - 3 = 9(b - 3)
5b - 3 = 9b - 27
5b - 9b = -27 + 3
-4b = -24
   b = -24/-4
   b = 6
Her son, Bob, is 6 years old

To find Agatha's age, substitute the value of b to the first equation
a = 5b
a = 5 × 6
a = 30
Agatha is 30 years old
4 0
3 years ago
Look at this cone:<br><br>,,
Assoli18 [71]
  • Slant height=l=3ft
  • Radius=r=2ft

We know

\boxed{\sf \star TSA_{(Cone)}=\pi r(r+\ell)}

\\ \sf\longmapsto TSA_{(Old\:Cone)}=2 \dfrac{22}{7}\times 2(2+3)

\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{44}{7}(5)

\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{220}{7}

\\ \sf\longmapsto TSA_{(Old\:Cone)}=31.4ft^2

Now

  • New slant height =2(3)=6cm
  • New radius=2(2)=4cm

\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{22}{7}\times 4(4+6)

\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{88}{7}(10)

\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{880}{7}

\\ \sf\longmapsto TSA_{(New\:Cone)}=125.7cm^2

So

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{125.7}{31.4}

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{4}{1}

\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cone)}}:{TSA_{(Old\:Cone)}}=4:1}}}

7 0
3 years ago
Please please please help me with two questions. Answer correctly for brainlist!!!!
Vesnalui [34]
1. 28.57  

2.  16
 
Hope this is right. 
5 0
3 years ago
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