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3 years ago

14

Judson collected information about the name length and the population of a random sample of 296 American cities. Here are the re

sults:
Name length Less than Between 150,000 More than Total
150,000 and 250,000 250,000
Less than 8
characters 30 29 41 100
8 to 10
characters 43 41 33 117
More than 10
characters 26 28 25 79
Total 99 98 99 296
Judson wants to perform a X2 test of independence between name length and population. What is the expected count for the cell corresponding to cities with more than 250,000 residents whose name has less than 8 characters?

1 answer:

Natasha2012 [34]3 years ago

6 0

It would be 90 - 10

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Someone help me please

natali 33 [55]

Part A:
1) 16
2) -25
3) 9
4) -7

I'm too lazy to do the rest, just pay attention in class next time

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3 years ago

You find 13,406,190 pennies, how many dollars doid you actually find? if each penny weighs 4 grams how much did all that loot we

RSB [31]

Answer: 134,061.9 dollar and 118,222.45 lb

To answer this question you need to convert each unit.
First, 1 dollar equal to 100 pennies. That mean the equation would be: <span>13,406,190 pennies x 1 dollar/100 pennies = 134,061.9 dollar

Each penny weight 4 grams and 1 lb equal to </span>453.592grams. Then the equation would be:
13,406,190 pennies x 4 grams/pennies x 1lb/453.592 grams= 118,222.45 lb

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3 years ago

Scale factor? i dont know im just trying to help my boyfriend within 6 minutes

iris [78.8K]

12.8 / 8 = 1.6
8 / 5 = 1.6

SF = 1.6
so x = 6 * 1.6 = 9.6

answer
x = 9.6

4 0

3 years ago

Match the parabolas represented by the equations with their foci.

Elenna [48]

Function 1 $f(x)=- x^{2} +4x+8$

First step: Finding when $f(x)$ is minimum/maximum
The function has a negative value $x^{2}$ hence the $f(x)$ has a maximum value which happens when $x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2$ . The foci of this parabola lies on $x=2$ .

Second step: Find the value of y-coordinate by substituting $x=2$ into $f(x)$ which give $y=- (2)^{2} +4(2)+8=12$

Third step: Find the distance of the foci from the y-coordinate
$y=- x^{2} +4x+8$ - Multiply all term by -1 to get a positive $x^{2}$
$-y= x^{2} -4x-8$ - then manipulate the constant of y to get a multiply of 4
$4(- \frac{1}{4})y= x^{2} -4x-8$
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is $12- \frac{1}{4}=11.75$

Hence the coordinate of the foci is (2, 11.75)

Function 2: $f(x)= 2x^{2}+16x+18$

The function has a positive $x^{2}$ so it has a minimum

First step - $x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4$
Second step - $y=2(-4)^{2}+16(-4)+18=-14$
Third step - Manipulating $f(x)$ to leave $x^{2}$ with constant of 1
$y=2 x^{2} +16x+18$ - Divide all terms by 2
$\frac{1}{2}y= x^{2} +8x+9$ - Manipulate the constant of y to get a multiply of 4
$4( \frac{1}{8}y= x^{2} +8x+9$

So the distance of focus from y-coordinate is $\frac{1}{8}$ to the north of $y=-14$
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: $f(x)=-2 x^{2} +5x+14$

First step: the function's maximum value happens when $x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25$
Second step: $y=-2(1.25)^{2}+5(1.25)+14=17.125$
Third step: Manipulating $f(x)$
$y=-2 x^{2} +5x+14$ - Divide all terms by -2
$-2y= x^{2} -2.5x-7$ - Manipulate coefficient of y to get a multiply of 4
$4(- \frac{1}{8})y= x^{2} -2.5x-7$
So the distance of the foci from the y-coordinate is - $\frac{1}{8}$ south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when $x=8.5$ and $y=79.25$ . The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when $x=-2.75$ and $y=-10.125$ . Manipulating coefficient of y, the distance of foci from y-coordinate is $\frac{1}{8}$ to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when $x=1.5$ and $y=9.5$ . The distance of the foci from the y-coordinate is $\frac{1}{8}$ to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

8 0

3 years ago

the aquarium has 30 times as many fish in one tank as jacob had. the aquarium has 90 fish. how many fish does jacob have

Stolb23 [73]

Jacob has three fish.

8 0

3 years ago