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alex41 [277]
3 years ago
12

PlEASE HELP ME PEOPLE HUMPTH OK ILL SAY IT PEOPLE WANT MONEY TO ANSWER PROBLEMS HERE If the mean of six nu m b e rs is 48, does

one
of the nu m b e rs have to be 48? Explain
why or why not. Give an example with six
nu m b e rs to show your answe
Mathematics
1 answer:
SpyIntel [72]3 years ago
6 0

Answer:

No, none of the number need to be 48 for the mean to be 48. To get a mean, you add up all the number and divide it by the amount of numbers.

Example:

the mean of 10, 79, 42, 88, 19, and 50 is 48, but the actual number 48 was not part of the set.

10 + 79 + 42 + 88 + 19 + 50 = 288

288 ÷ 6 = 48

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Answer:

each student gets 3 drinks, make sure the lemon and lime are in different groupings

Step-by-step explanation:

7 0
2 years ago
F(x)=12x2-4x-8. G(x)=11x-6
AlekseyPX

Answer: Second option.

Step-by-step explanation:

Given the functions f(x) and g(x):

f(x)=12x^2-4x-8\\\\g(x)=11x-6

You need to divide them in order to find (\frac{f}{g})(x) asked in the exercise. Then:

(\frac{f}{g})(x) =\frac{12x^2-4x-8}{11x-6}

Now, it is important to remember that, by definition, the division by zero is not defined. Therefore, the denominator of the function cannot be zero.

Let's find the value of "x" for which the denominator of the function would be zero:

1. You need to make the denominator equal to zero:

11x-6=0

2. Finally, you must solve for "x":

11x=6\\\\x=\frac{6}{11}

Therefore, as you can see, the answer is:

 (\frac{f}{g})(x) =\frac{12x^2-4x-8}{11x-6}, x=\frac{6}{11}

4 0
3 years ago
If andre collects $9.80, how many cups did he sell?
barxatty [35]

Answer:

we need more information

5 0
2 years ago
The formula for a trapezoid relates the area A, the two bases, a and b, and the height, h. A=(a+b)/2 h Solve for a.
yKpoI14uk [10]

Answer:

a=\frac{2A}{h}-b

Step-by-step explanation:

Hello, I think I can help you with this.

According to the information provided in the problem, the area of ​​a trapezoid is given by

A=\frac{a+b}{2} *h\\

Step 1

solve for a, it means isolate a

A=\frac{a+b}{2} *h\\divide\ both\ sides\ by\ h\\\\\frac{A}{h}= \frac{\frac{a+b}{2} *h}{h} \\\frac{A}{h}=\frac{a+b}{2}\\\\Now, multiply\ both\ sides\ by\ two\\\\ 2*\frac{A}{h}=2*\frac{a+b}{2}\\2*\frac{A}{h}=a+b\\\frac{2A}{h}=a+b\\\\subtract\ b\ for\ both\ sides\\\frac{2A}{h}-b=a+b-b\\a=\frac{2A}{h}-b

Have a good day.

5 0
3 years ago
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