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Murrr4er [49]
3 years ago
6

PLEASE ANSWER ASAP! 15 POINTS ! NO SCAM LINKS OR REPORTED :)

Mathematics
1 answer:
mixas84 [53]3 years ago
5 0

Answer: C

Step-by-step explanation: You have to plug in the values to see if it matches the values on the table of values

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How do you distribute to solve the equation 28-2(4^2*4)?
astra-53 [7]

Answer:

28-2(4^2*4)

28 - 2 (4*4*4)

28-2(64)

28 -128 = 100

Step-by-step explanation:

28-2(4^2*4)

28 - 2 (4*4*4)

28-2(64)

28 -128 = 100

5 0
4 years ago
Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)f(x) =
lyudmila [28]

Answer:

The absolute minimum value is "-\frac{21}{4}" and the absolute maximum value is "15".

Step-by-step explanation:

Given:

f(x)=x^2-x-5

on,

[0,5]

By differentiating it, we get

⇒ f'(x)=2x-1

Set f'(x)=0

then,

⇒ 2x-1=0

          2x=1

            x=\frac{1}{2} (Critical point)

When x=0,

⇒ f(x)=-5

When x=\frac{1}{2},

⇒ f(x)=-\frac{21}{4} (Absolute minimum)

When x=5

⇒ f(x)=15 (Absolute maximum)

7 0
3 years ago
The value of a company's stock decreases by $15,000. If each share decreased in value from $8 1/16 to $8.00 how many shares are
lina2011 [118]
There are 1875 shares in the company i believe

4 0
3 years ago
Let x1 = 12, y1 = 15, and y2 = 3. Let y vary inversely with x. Find x2.
Ostrovityanka [42]
We have that an inverse variation in a function is given by:
 y = k / x

 We must find the value of the constant k.
 To do this, we substitute the values of x1 = 12, y1 = 15.
 We have then:
 15 = k / 12

 Clearing k we have:
 k = (15) * (12)

k = 180
 Substituting the value of k we have that the inverse variation is:
 y = 180 / x

 For y2 = 3 we have:
 3 = 180 / x2

 Clearing x2 we have:
 x2 = 180/3
 x2 = 60
 Answer:
 
x2 = 60
3 0
4 years ago
To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the
Lisa [10]

Solution :

Group   Before     After

Mean    693.75    743.75

Sd         155.37     143.92

SEM        54.93     50.88

n              8            8

Null hypothesis : The preparation course not effective.

$H_0: \mu_d = 0$

Alternative hypothesis : The preparation course is effective in improving the exam scores.

$H_a : \mu_d>0$  (after -  before)

4 0
3 years ago
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