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Vesna [10]
3 years ago
11

Please Help Soon. I need this by Wednesday

Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0
Answer : b = 81/4 = 20.25
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Hey can you please help me posted picture of question
dangina [55]
The solutions to the given quadratic equation can be determined using quadratic formula.

x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a}

b =coefficient of x term = -1
a = coefficient of squared term = 4
c = constant term = 9

Using the values in the above formula we get:

x= \frac{1+- \sqrt{1-4(4)(9)} }{2(4)}  \\  \\ 
x= \frac{1+- \sqrt{-143} }{8}  \\  \\ 
x= \frac{1+-i \sqrt{143} }{8}

So, option D gives the correct answers


4 0
3 years ago
30. If the level of significance is 0.05 and the p-value is 0.04, what conclusion can you draw?
Inessa [10]

Answer:

Reject null hypothesis

Step-by-step explanation:

Because our p-value of 0.04 is less than the level of significance of 0.05, then we reject the null hypothesis that there's no difference between the means and conclude that a significant difference does exist

4 0
3 years ago
Giant is selling Cereal at $8 for 3 boxes.what is the cost per box?
Zolol [24]

Answer:

$2.67

Step-by-step explanation:

you divide to find unit rate

8/3 = 2.666667

7 0
3 years ago
Read 2 more answers
Sophia took a test with 21 questions. If she answered two sevenths of the questions​ correctly, how many did she get​ wrong?
guajiro [1.7K]

Answer:

Sophia took a test with 21 questions.

If she answered two sevenths of the questions​ correctly, number of correct answers she did: N = 21 x 2/7 = 6

=> Number of answers she got wrong: N = 21 - 6 = 15

Hope this helps!

:)

4 0
3 years ago
Read 2 more answers
If f(x) = 1/3x+13, then f^-1(x) =??​
Sergeeva-Olga [200]

Step-by-step explanation:

Given

f(x) = 1/3x + 13

f^-1(x) = ?

Now

Let y = f(x)

or y = 1/3x + 13

Interchanging the role of x and y

x = 1/3y + 13

x - 13 = 1/3y

y = <u>x </u><u>-</u><u> </u><u>1</u><u>3</u>

3

Therefore f^-1(x) = <u>x </u><u>-</u><u> </u><u>1</u><u>3</u><u> </u>

3

Hope it will help you.

7 0
3 years ago
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