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3 years ago

30 POINTS! PLZ HELPPP!!! The list shows the number of children who visited a city playground each day for two weeks.

1 answer:

5 0

Answer to first part 1: First you obtain number of intervals you would like to have by getting the square root of the total number of observations. In this case, square root of 14 = 3.74; round up and use 4.0.
Answer to Second part: 2: The interval width is then obtained by dividing the range by the number of intervals. For this case, interval width = (42-9)/4 = 8.25. This can be rounded up to the convenient evenfigure of 10.0.

Hope this helps!!!!

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Anyone please help!!!

Tanya [424]

Problem 1

Answers:

Percentage of patients that were dogs = 46%

Standard Error = 0.07048404074682

Margin of error for 90% confidence interval = 0.11594624702851

Margin of error for 95% confidence interval = 0.13814871986377

Round the decimal values however you need them

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To get the first answer, you add up the numbers given (7,4,5,5,2) and divide that over 50. So 7+4+5+5+2 = 23 which leads to 23/50 = 0.46 = 46%; therefore phat = 0.46 is the sample proportion of dogs.

Use the SE (standard error) formula given to you with phat = 0.46 and n = 50 to get SE = 0.07048404074682

The critical z value at 90% confidence is 1.645; this value is found in your Z table (back of your stats textbook). Multiply the SE value by 1.645 to get 0.07048404074682*1.645 = 0.11594624702851

Also found in your textbook is 1.960 which is the z critical value at 95% confidence. Multiply this with the SE value to get 0.07048404074682*1.960 = 0.13814871986377

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Problem 2

Answer: Choice B) picking balls from a bin; the 60 randomly selected get chosen for the first bus, while the remaining 60 go to the second bus

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Choice A is fairly vague on what the lower and upper boundaries are. What is the smallest number allowed? What about the largest? This isn't clear so it's possible that we could end up with more positive numbers than negative (eg: if we had an interval -10 < x < 110). So choice A is false. A similar issue shows up with choice D.

Choice B is true. Assuming the selection process is random and not biased, then each ball is equally likely for each selection. The fact that the balls are colored seems to be extra info which I'm not sure why your teacher threw that in there.

Choice C is false because choice B is true

Choice D is false for similar reasons as choice A. It's not clear where we start and where we end. If we had the interval 2 < x < 6 then x could take on the values {3, 4, 5} and we see that picking an odd number is twice as likely than picking an even. In this example, there is bias.

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Problem 3

Answer: Choice B) Roll a die; each number corresponds to a different class

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Choice A is false because choice B being the answer contradicts it

Choice B is true: there are 6 sides on the die, and each side is equally likely to be landed on, so each class is equally likely

Choice C is false because 2*2*2 = 8 represents the number of combos you can have when you flip three coins (one combo being HTH for heads tail heads) but there are 6 classes, not 8

Choice D is false because while we want 6 regions on the spinner. Each region must have the same area; otherwise, one class is weighted heavier than the others making it more likely you select that particular class.

7 0

3 years ago

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In the adjoining figure, XY = XZ . YQ and ZP are the bisectors of <img src="https://tex.z-dn.net/?f=%20%5Cangle" id="TexFormula1

svetlana [45]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$1)\, XY=XZ$ Given

$2) \text{ $ m\angle Y= m\angle Z$}$ Isosceles Triangle Theorem

$\displaystyle 3) \text{ $m\angle Y=m\angle XYQ + \angle QYZ$}$ Angle Addition

$\displaystyle 4)\text{ $YQ$ bisects $\angle XYZ$}$ Given

$5) \text{ $m\angle XYQ=m\angle QYZ$}$ Definition of Bisector

$\displaystyle 6)\text{ $m\angle Y=2m\angle QYZ$}$ Substitution

$7)\text{ $m\angle Z=m\angle XZP+m\angle PZY$}$ Angle Addition

$8)\text{ $ZP$ bisects $\angle XZY$}$ Given

$\displaystyle 9) \text{ $m\angle XZP=m\angle PZY$ }$ Definition of Bisector

$\displaystyle 10) \text{ $ m\angle Z = 2m\angle PZY $}$ Substitution

$11)\text{ } 2m\angle QYZ=2m\angle PZY$ Substitution

$12)\text{ }m\angle QYZ=m\angle PZY$ Division Property of Equality

$13)\text{ } YZ=YZ$ Reflexive Property

$14)\text{ } \Delta YZP\cong\Delta ZYQ$ Angle-Side-Angle Congruence*

$15)\text{ } YQ=ZP$ CPCTC

*For clarification:

∠Y = ∠Z

YZ = YZ (or ZY)

∠PZY = ∠QYZ

So, Angle-Side-Angle Congruence:

ΔYZP is congruent to ΔZYQ

5 0

3 years ago

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R-12 evaluate if r= negative 3

GalinKa [24]

If r=-3, then it would be -3-12 which equals -15

5 0

4 years ago

PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

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

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

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

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

7 0

3 years ago

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5. Janna and Laura broke a six-inch candy bar in half to share with each

kobusy [5.1K]

Answer:

3 inches

Step-by-step explanation:

6 divded by 2 (or in half ) is 3

4 0

3 years ago

Read 2 more answers