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SOVA2 [1]
3 years ago
6

Find the area plzzzzzzxxxzzxxxzzzzzzz

Mathematics
1 answer:
nikitadnepr [17]3 years ago
5 0

Step-by-step explanation:

step 1. break up the shape into 2 triangles and one rectangle

step 2. A = area of triangle 1 (t1) + area of triangle 2 (t2) + area of the rectangle (r)

step 3. t1 = (1/2)bh where b is the base and h is the height. definition of the area of a triangle.

step 4. t1 = (1/2)(3)(2) = 3ft^2.

step 5. t2 = (1/2)(3)(3) = 4.5ft^2.

step 6. r = bh where b is the base and h is the height. definition of the area of a rectangle.

step 7. r = bh = (8)(4) = 32ft^2.

step 8. A = t1 + t2 + r = 3 + 4.5 + 32 = 39.5ft^2.

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What is 1/6 of 54= 1/3 of WHAT
hodyreva [135]

Answer:

27

Step-by-step explanation:

⅙ × 54 = ⅓ × x

9 = x/3

x = 27

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3 years ago
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Can someone please help me, someone who's serious lol. PLEASE
klemol [59]

Question 11.4.10

Add up everything in the first row

13+31+24 = 68

We have 68 people with high school only education. Out of this group, 24 are conservative

So we have 24/68 = 6/17 represent the probability of selecting a conservatives person given they have high school only education.

<h3>Answer:  6/17</h3>

=======================================================

Question 11.4.15

There are 269+485 = 754 males.

Of this group, 485 are not associates.

Therefore, 485/754 = 0.64 approximately

We focus on the males only because of the phrasing "given that the recipient is male".

<h3>Answer:  0.64 </h3>

=======================================================

Question 11.4.16

Refer to the diagram below. Let's say we had 1000 people surveyed. This means we have 1000 in the bottom right corner to represent the grand total.

40% have a pet now currently and also had a pet in the past, so we have 0.40*1000 = 400 people go in the upper left corner of the table. This is the "have a pet now" row and "had a pet in the past" column.

60% do not have a pet now, so 600 people don't have a pet currently. This is the total for row 2. This must mean 1000-600 = 400 people do have a pet currently. This is the total for row 1.

The missing value in row 1 must be 0 because 400+0 = 400. Also, if a person has a pet now, they likely had it in the past as well. To be honest, the wording to set up the time scale for this problem is vague. What does it mean "in the past"? Do the researchers mean in the previous month? Year? It's not clear. Anyway, the first row consists of 400, 0 and 400 in that order.

----------------------

Then we're told that 85% have had a pet in the past. So 0.85*1000 = 850 people had a pet in the past. This goes at the bottom of the "had a pet in the past" column. This leaves 1000-850 = 150 people who didn't have a pet in the past.

Lastly, 15% of the 1000 people surveyed don't have a pet in the present and they didn't have it in the past either. Since 15% of 1000 = 0.15*1000 = 150, we will write 150 in the second entry of row 2. Notice how 0+150 = 150 for the second column. Every row should add up to the entry in the "total" column, and every column should add up to the entry in the "total" row to help check we have the correct table set up. Lastly, notice how 400+600 = 1000 and 850+150 = 1000 to further verify the table.

----------------------

By this point, the table is completely filled out. We're told "given the respondent had a pet". We only focus on the "had a pet in the past" column. 400 people have a pet now out of 850 who had a pet in the past.

Therefore, 400/850 = 0.47 approximately

<h3>Answer: 0.47</h3>

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RIGHT ANSWER WILL BE MARK BRAINLIEST
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What is breaking probation?
masha68 [24]

Answer:

a probation violation occurs when someone who has already been sentenced for a crime breaks the terms or conditions of that probation

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PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
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