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igor_vitrenko [27]
4 years ago
9

50, 45, 49, 50, 43, 49, 50, 49, 45, 49, 47, 47, 44, 51, 51, 44, 47, 46, 50, 44, 51, 49, 43, 43, 49, 45, 46, 45, 51, 46.what is t

he range for the data set shown
Mathematics
1 answer:
erma4kov [3.2K]4 years ago
7 0
The range is 7 because the highest number is 50 and the lowest is 43. 
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Solve the equation: Ab = 56 b = [?]​
riadik2000 [5.3K]

Answer:

B can be 1, 2, 4, 7, 8, 14, 28 and 56.

Basically any factor of 56.

Hope that helps!

<em>-scsb17hm</em>

Step-by-step explanation:

7 0
3 years ago
Solve negative one fifth times the quantity 3 minus 12 times the quantity one third squared end quantity end quantity.
katrin [286]

The solution to the given equation is -87/45

<h3>Product of expressions</h3>

Given the following expression that represents the word problem as;

-1/5 * 3 - 12(1/3)^2

Using the PEMDAS rule, the parenthesis and exponent will come first to have;

-1/5 * 3 - 12(1/3)^2 = -1/5 * 3 - 12(1/9)

Then multiplication

-1/5 * 3 - 12(1/9) =. -3/5 - 12/9

Find the difference

-1/5 * 3 - 12(1/9) = -27-60/45

-1/5 * 3 - 12(1/9) = -87/45

Hence the solution to the given equation is -87/45

Learn more on PEMDAS here; brainly.com/question/345677

#SPJ1

4 0
2 years ago
Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x &gt; 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
4 years ago
Read 2 more answers
-3/4+p=1/2 anserwe this plz thanks
mrs_skeptik [129]

Answer: p=5/4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
12. Find the values of x and y in the diagram below
Lapatulllka [165]

Answer:

1. y = 18/4

2. x = 11/6

3. y = -14

4. x = -5

Step-by-step explanation:

1. 4y - 18 = 0

4y = 18

y = 18/4

2. 6x - 11 = 0

6x = 11

x = 11/6

3. y + 14 = 0

y = -14

4. x + 5 = 0

x = -5

4 0
4 years ago
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