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Mademuasel [1]
2 years ago
13

when asher looks at the data he says that both Mrs. Jamisos class and Mr. Zimmermann's class has data that is skewed rigjt. Do y

ou agree ir disagree
Mathematics
1 answer:
pentagon [3]2 years ago
6 0

Answer:

can you please give more info

Step-by-step explanation:

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Sam would have 41 pence change
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What is 3852 divided by 16
Annette [7]
240.75, hope this helps!
3 0
3 years ago
The value of x satisfying both the equations 4x – 5 = y and 2x – y = 3, when y = -1 is
schepotkina [342]

Answer:

x = 1

Step-by-step explanation:

4x-5=y

4x-5= -1

4x = -1+5

4x=4

x=1

2x-y=3

2x-(-1)=3

2x+1=3

2x= 3-1

2x= 2

x=1

since the two answers for x are the same therefore:

x = 1

4 0
3 years ago
Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f
quester [9]

Answer:

It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x) (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

5 0
3 years ago
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Whats the length of the hypotenuse of a right triangle if the length of one leg is 7 units and the length of the other leg is 14
Artyom0805 [142]

Answer:

The answer is D ( 15.65 )

Step-by-step explanation:

I did it (:

8 0
3 years ago
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