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irga5000 [103]
3 years ago
6

Use the monthly payment formula to find the monthly payment for a $1,000, one year

Mathematics
1 answer:
wlad13 [49]3 years ago
6 0
<h3>Answer:  89,58 $ </h3>

Step-by-step explanation:

The formula :

\displaystyle \rm S=A\cdot \left (1+\frac{N}{100}  \right  )^{\big r}

where r is years ; N is the percentage by which we increase the price ; A is the original price

In our case :

\rm r=1 \ \ ;  \ \  A=100 \ \ ;   \  \  N=7,5\% \\\\ S=1000\cdot \bigg(  1+\dfrac{7,5}{100} \bigg)^1=1000\cdot 1,075=\boxed{1075\$}

Then the monthly salary is equal to:

\dfrac{1075}{12} \approx 89,58 \$

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(+×9)(×-2) can some one solve it
marshall27 [118]
X2 - 2x + 9x - 18
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4 0
3 years ago
Help pleaseeeeeeeeee
sergij07 [2.7K]

9514 1404 393

Answer:

  47 -6√10

Step-by-step explanation:

As you know, the area of a square is the square of the side length. It can be helpful here to make use of the form for the square of a binomial.

  (a -b)² = a² - 2ab + b²

  (√2 -3√5)² = (√2)² - 2(√2)(3√5) + (3√5)²

  = 2 - 6√10 + 3²(5)

  = 47 -6√10

__

<em>Check</em>

  √2-3√5 ≈ -5.29399 . . . . . . . . note that a negative value for side length makes no sense, so this isn't about geometry, it's about binomials and radicals

  (√2-3√5)² ≈ 28.02633

  47 -6√10 ≈ 28.02633

5 0
3 years ago
The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir
Paha777 [63]

Answer:

95​% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                              P.Q. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean time = 5.15 years

            \sigma = sample standard deviation = 1.68 years

            n = sample of college graduates = 4400

            \mu = population mean time

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                               level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                              = [ 5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } } , 5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } } ]

                                             = [5.10 , 5.20]

Therefore, 95​% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0
3 years ago
Find f.<br><br> A) 7.4<br><br> B) 8.2<br><br> C) 10.5<br><br> D) 11.1
tatyana61 [14]
F=72

g=6

------------

\cos { \left( F \right)  } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }

Therefore:

\cos { \left( 72 \right)  } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right)  } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }

\\ \\ 12e\cdot \cos { \left( 72 \right)  } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right)  } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 }  } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right)  }  \right) +36 }

But what is e?

E=76

G=32

g=6

And:

\frac { e }{ \sin { \left( E \right)  }  } =\frac { g }{ \sin { \left( G \right)  }  }

Which means that:

\frac { e }{ \sin { \left( 76 \right)  }  } =\frac { 6 }{ \sin { \left( 32 \right)  }  } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  }

If you take this value into account, you will discover that f is...

f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  } \left( \frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  } -12\cos { \left( 72 \right)  }  \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)

So I would have to say that the answer is approximately (c).
6 0
3 years ago
What is the number property of 17+0
vladimir1956 [14]

Answer:

The identity property

Step-by-step explanation:

anything added to 0 is itself.

8 0
3 years ago
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