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Alexeev081 [22]
3 years ago
7

PLEASE HELP, if u could look at my answer it says 48 inches is that correct if it's not could you tell me the right answer.​

Mathematics
1 answer:
andreev551 [17]3 years ago
4 0
Yes I believe it is correct but not sure
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8 – 6 ∙ 4 + 10 ÷ 2 =
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Answer:

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What is "h"? h+1/2=3 1/4
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For this equation
H+1/2= 3 1/4

H is a variable, so it must have a value to it. So it would be 2.75+1/2= 3 1/4
Or 3 1/4 / 1/2 = 2.75

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The solution to a quadric equation are x=2 and x=7. If k is a nonzero constant which of the following must be equal to the quadr
nikklg [1K]

Answer:

The quadratic equation form is x² - 9 x + 14  = 0            

Step-by-step explanation:

Given in the question as ,

The solution of quadratic equation are x = 2   and x = 7

The constant K is a non-zero number

Now , let the quadratic equation be , ax² + bx + c = 0

And  sum of roots = \frac{ - b}{a}

       product of roots = \frac{c}{a}

So, 2 + 7 = \frac{ - b}{a}

Or, 2 × 7 =  \frac{c}{a}

I.e \frac{ - b}{a}  = 9     ,     \frac{c}{a} = 14

Or,  b =  - 9 a           And c = 14 a

Put this value of b and c in standard form of quadratic equation

I.e ax² + bx + c = 0

Or, ax² - 9 ax + 14 a = 0

Or , a ( x² - 9 x + 14 ) = 0  

∴ a = 0   And  x² - 9 x + 14 = 0

Hence ,  The quadratic equation form is x² - 9 x + 14  = 0      Answer

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3 years ago
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almond37 [142]

the anwser is ]123[8D[

7 0
3 years ago
Exercise 1.A committee of five people is chosen randomly from four men and six women. Find the probability that:a) Exactly four
MatroZZZ [7]

Answer with explanation:

Given : Number of men = 4

Number of women = 6

Total people = 6+4=10

Total number of ways to make a committee of five people from 10 persons :-

^{10}C_{4}=\dfrac{10!}{4!(10-4)!}=210

a) Number of ways to make committee that has exactly four women :

^6C_4\times ^4C_1=\dfrac{6!}{4!(6-4)!}\times4=60

The  probability that committee has exactly four women :

\dfrac{60}{210}=\dfrac{2}{7}

b) Number of ways to make committee that has at-least four women :

^6C_4\times ^4C_1+^6C_5=\dfrac{6!}{4!(6-4)!}\times4+6=66

The probability that committee at-least four women :

\dfrac{66}{210}=\dfrac{22}{70}

c) Number of ways that committee has more than 4 women :-

^6C_5\times^4C_0=6

The probability that committee has more than 4 women :-

\dfrac{6}{210}

Now, the  probability that committee has at most four women :-

1-\dfrac{6}{210}=\dfrac{204}{210}=\dfrac{102}{105}

5 0
3 years ago
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