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erastova [34]
3 years ago
7

Nate solve the equation 3 times 5 equals 15 by writing and solving 15 divided by 5 is 3 x plane why Nate's method works

Mathematics
1 answer:
Vsevolod [243]3 years ago
8 0
Division is the opposite from multiplication
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How to do this and I need help pls
tatuchka [14]

Answer:

9

Step-by-step explanation:

f(x) = (x - 2)^{2}  \\ f(5) = (5 - 2)^{2}  \\ f(5) =  {(3)}^{2} \\ f(5) = 3 \times 3 \\ f(5) = 9

7 0
3 years ago
The wall street journal corporate perceptions study 2011 surveyed readers and asked how each rated the quality of management and
malfutka [58]

Answer:

Hello some parts of your question is missing below is the missing part

answer : A) p-value =  0.0019 ,

The level of significance ( 0.05 ) > p-value ( 0.0019 )

B) There is no dependence between the two ratings

Step-by-step explanation:

A) using a 0.05 level of significance and test for independence of the quality of management

Test statistic : X^{2}  ∑ ( Oi - Ei )^2 / Ei

= [(40-35)^2 / 35 ] +  [(25-24.5)^2 / 24.5 ] +  [(5-10.5)^2 / 10.5]  +  [(35-40)^2 / 40]  +  [(35-28)^2 / 28]  +  [(10-12)^2 / 12]  +   [ (25-25)^2 / 25 ]  +  [(10-17.5)^2 / 17.5 ]  + [(15-7.5)^2 / 7.5]  =  17.03  

p-value = 0.0019

Df = 4  ,

The level of significance ( 0.05 ) > p-value ( 0.0019 )

B) There is no dependence between the two ratings

4 0
2 years ago
For a school play, 739 tickets valued at $857 were sold. Some cost $1 and others cost a $1.50. How many $1 tickets were sold?
balu736 [363]

Answer:

503 $1 tickets sold.

Step-by-step explanation:

Use two equations  

Let x = number of $1 tickets sold  

Let y = number of $1.50 tickets sold  

x + y = 739  

1x + (1.5)y = 857  

First equation ==> y = 739 - x  

Plug this into the second equation  

x + (1.5)(739 - x) = 857  

x + 1108.5 - 1.5x = 857  

- 0.5x = -251.5  

x = 503  

There were 503 $1 tickets sold.  

To find the number of $1.50 tickets, just plug this value of x into either one of the equations.  

(503) + y = 739       (739 - 503 = 236)

y = 236  

There were 236 $1.50 tickets sold.

3 0
3 years ago
1 point) Use Stoke's Theorem to evaluate ∫CF⋅dr∫CF⋅dr where F(x,y,z)=xi+yj+1(x2+y2)kF(x,y,z)=xi+yj+1(x2+y2)k and CC is the bound
Drupady [299]

Answer:

0

Step-by-step explanation:

Thinking process:

\int\limits^a_b {cF} \, .dr

= \int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx curlFdS by Stoke's Theorem

= \int\limits^a_b {} \,  \int\limits^a_b {} \, < 12y, -12x, 0 > .< z_x,-z_y, 1 > dA\\= \int\limits^a_b {} \, \int\limits^a_b {} \, < 12y, -12x, 0 > .< 2x, 2y, 1 > dA since z = 25-x^{2} -y^{2} \\= 0

3 0
3 years ago
What is the answer and how do you solve <br><br><br> 21x^3/3x^−1
Sunny_sXe [5.5K]

Answer:

  as written: 7x^2

  perhaps intended: 7x^4

Step-by-step explanation:

As written, the Order of Operations specifies that the only denominator factor is 3, so the expression is interpreted as ...

  \dfrac{21x^3}{3}x^{-1}=\dfrac{21}{3}x^{3-1}=7x^2

___

Sometimes grouping of denominator factors is intended, but parentheses are not written as they should be. If the intention is ...

  21x^3/(3x^-1)

then the simplification is ...

  \dfrac{21x^3}{3x^{-1}}=\dfrac{21}{3}x^{3-(-1)}=7x^4

_____

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  a^b/a^c = a^(b-c)

6 0
3 years ago
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