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Aleksandr [31]
3 years ago
8

Which of the following statements is correct if you roll a fair 6-sided die 600 times?

Mathematics
1 answer:
never [62]3 years ago
7 0
The answer is b hope it helped
You might be interested in
Okay, what is x if: -7/9 (x 3) =14?
Charra [1.4K]
I hope this helps you



-7.x^3=2.7.9



x^3=-18


x=i.(3square root of 18)
3 0
2 years ago
Simplify.<br> -12 + (-2)(3)
sammy [17]

Answer:

-18

Step-by-step explanation:

-12 + (-2)(3)

= (-12)+((-2)x(3))

= (-12)+(-6)

= (-12)-(6)

= -18

8 0
3 years ago
Hi! ❤️ , im looking for some help here. ill give brainliest if able to.
Alexeev081 [22]

Answer:

A. 2^11

Step-by-step explanation:

(They are basically asking what's 2^4 × 2^7, but with more words.)

I usually do each exponent individually:

2^4 is the same as 2 × 2 × 2 × 2 = 16 (or you could have read the text to figure that out)

2^7 is the same as 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128

Then just multiply 128 and 16 to get 2,048, and see which option also gives you 2,048.

BUT, you can also:

(Combine the exponents together to get your answer. Just remember that if it's multiplication you add them, and if it's division you subtract them.)

2^4 × 2^7

4 + 7 = 11

2^11 (This equals 2,048 btw. You don't even have to check all the options to get the answer).

Hope this helps friend :)

The last part I learned from another user, while answering one of your other questions. I personally find this mind blowing, lol.

4 0
2 years ago
WILL GIVE BRAINLIEST AND 70 POINT! PLEASE HELP ME!
PilotLPTM [1.2K]

Answer:

X and Y

and for ast two answers pls give me the options

7 0
3 years ago
Read 2 more answers
The quadratic formula gives which roots for the equation 2x^2 + x - 6 = 0?
Jet001 [13]

Answer:

The roots are x = (\frac{3}{2}, -2), which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}

x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}

\Delta = b^{2} - 4ac

In this question, we have that:

2x^2 + x - 6 = 0

Which is a quadratic equation with a = 2, b = 1, c = -6. So

\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49

x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}

x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2

So the roots are x = (\frac{3}{2}, -2), which is given by option C.

3 0
3 years ago
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