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3 years ago

What is the equation for 1.2×5 in fractions

Mathematics

2 answers:

creativ13 [48]3 years ago

4 0

I think six...........

mariarad [96]3 years ago

3 0

1.2 times 5 equals 6 since there’s a decimal involved you have to move it so it would be 0.6 but sincere it’s fractions it would be 6 over 10

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Find the root(s) of the equation x2 = 121?

Andru [333]

±11

Step-by-step explanation:

Even NO. roots are always contain ±.

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2 years ago

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What is the answer to the question?

ololo11 [35]

Answer:

I think the answer is -25

Step-by-step explanation:

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3 years ago

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Identify the zero(s) of <img src="https://tex.z-dn.net/?f=f%28x%29%3D4x%5E%7B2%7D%20-%2020x%20-%2056" id="TexFormula1" title="f(

AURORKA [14]

Answer:

x = - 2, x = 7

Step-by-step explanation:

Given

f(x) = 4x² - 20x - 56

To find the zeros let f(x) = 0, that is

4x² - 20x - 56 = 0 ( divide through by 4 )

x² - 5x - 14 = 0 ← in standard form

(x - 7)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

x - 7 = 0 ⇒ x = 7

5 0

2 years ago

Two variables are correlated with r = -0.23. Which description best describes the strength and direction of the association betw

tatuchka [14]

Answer is C weak negavite

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3 0

3 years ago

A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang

laiz [17]

<u>Answer-</u>

a. Probability that three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

$\binom{19}{3} = \frac{19!}{3!16!} = 969$

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

$\binom{33}{6} = \frac{33!}{6!27!} =1107568$

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

$\binom{52}{9} = \frac{52!}{9!43!} = 3679075400$

∴P(3 white candies) = $\frac{1073233392}{3679075400} =0.29$

Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

$\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}$

$=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})$

$=(969)(45)(7)(5)(15)=22892625$

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

$\binom{52}{9}=\frac{52!}{9!43!} =3679075400$

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

$\frac{22892625}{3679075400} = 0.006$

6 0

3 years ago