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Ksivusya [100]
3 years ago
11

Find a 95% confidence interval for the mean auditory response time for subjects with a visual response time of 200 ms. The 95% c

onfidence interval for the mean auditory response time for subjects with a visual response time of 200 ms is:____.
Mathematics
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

hello your question is incomplete attached below is the complete question

answer : ( 184.019 , 206.219 )

Step-by-step explanation:

The 95% confidence interval for the mean auditory response time for subjects with a visual response time of 200 ms can be calculated using this relationship below

y = Bo + B1 ( 200 )  = 195.118621 , sy =

hence a 95% confidence interval = 195.118621 ± 2.306( 4.813492) i.e.  ( 184.019 , 206.219 )

attached below is the remaining part the solution

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A cleaning company charges x dollars per hour to clean floors and y dollars per hour to clean the rest of the house
Nina [5.8K]

Answer:

(15,18)

Step-by-step explanation:

write two equations with the information provided.

2x + 3y = 84

x + 4y = 87

use the property of substitution to answer.

x + 4y = 87,   x = 87 - 4y

2(87 - 4y) + 3y = 84

174 - 8y + 3y = 84

174 - 84 = 8y - 3y

90 = 5y

90/5 = y

y = 18

Add the value of Y to an original equation. Solve for X

2x + 3(18) = 84

2x + 54 = 84

2x = 84 - 54

2x = 30

x = 30/2

x = 15

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3 years ago
The spinner below is spun twice. What is the probability of the arrow landing on a 3 and then an odd number?
Mice21 [21]

Answer:

4/3

Step-by-step explanation:

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-38 due to the fact the lower the number, the colder it is.

4 0
3 years ago
Read 2 more answers
What is the answer ?
kifflom [539]
The system of inequalities is the following:

i) <span>y ≤ –0.75x
ii)</span><span>y ≤ 3x – 2

since </span>0.75= \frac{75}{100}= \frac{3}{4}, we can write the system again as 

i) y \leq - \frac{3}{4}x
ii) y  \leq 3x-2

Whenever we are asked to sketch the solution of a system of linear  inequalities, we:

1. Draw the lines
2. Color the regions of the inequalities.
3. The solution is the region colored twice.


A.

to draw the line y =- \frac{3}{4}x

consider the points: (-4, 1) and (0, 0), or any 2 other points (x,y) for which y =- \frac{3}{4}x hold.

since we have an "smaller or equal to" inequality, the line is a solid line (not dashed, or dotted).

In order to find out which region of the line to color, consider a point not on the line, for example P(1, 1), which is clearly in the upper region of the line.

For (x, y)=(1, 1) the inequality y  \leq - \frac{3}{4}x, does not hold because 

1 \leq - \frac{3}{4}*1= -\frac{3}{4} is not true,

this means that the solution is the region of the line not containing (1, 1), as shown in picture 1.


B.
similarly, to draw the solution of inequality ii) y ≤ 3x – 2, 

we first draw the line y=3x-2, using the points (0, -2) and (2, 4), or any other 2 points (x,y) for which y=3x-2 holds.

after we draw the line, we can check the point P(1, 7) which clearly is above the line y=3x-2.

for (x, y) = (1, 7), the inequality y ≤ 3x – 2 does not hold

because 7 is not ≤ 3*1-2=1, so the region we color is the one not containing P(1, 7), as shown in picture 2.


The solution of the system is the region colored with both colors, the solid lines included. Check picture 3.

the lines intersect at (0.533, -0.4) because:

–0.75x=3x-2
-0.75x-3x=-2
-3.75x=-2, that is x= -2/(-3.75)=0.533

for x=0.533, y=3x-2=3(0.533)-2=-0.4

Answer: Picture 3, the half-lines included. So the graph is in the 3rd and 4th Quadrants

8 0
3 years ago
Fiona is trying to prove that a pair of expressions are equivalent by using substitution. Which pair of expressions has the same
fiasKO [112]

Missing Part of the Question

  • 3 (4 - y) -7 ; 5 + 3y
  • 9 + y + 4y - 2 - 3y; -2y + 7
  • 5 + y - 2 - 6y + y; 8y + 3
  • 2y + 7 + y - 2 + y; 4y + 5

Answer:

  • 2y + 7 + y - 2 + y; 4y + 5

Step-by-step explanation:

Given

y = 3

Required

Which pair of expression are equal

To get the pair of expression that are equal; we simply substitute 3 for y in each of the expressions

3 (4 - y) -7 and 5 + 3y

<em>Substitute 3 for y in 3 (4 - y) -7</em>

This gives 3(4 - 3) -7

= 3(1) - 7

= 3 - 7

= -4

<em>Then Substitute 3 for y in 5 + 3y</em>

This gives  5 + 3(3)

= 5 + 9

= 14

-4 and 14 are not equal

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

9 + y + 4y - 2 - 3y; -2y + 7

<em>Substitute 3 for y in </em>9 + y + 4y - 2 - 3y

This gives 9 + 3 + 4(3) - 2 - 3(3)

= 9 + 3 + 12 - 2 - 9

= 13

<em>Then Substitute 3 for y in -2y + 7</em>

This gives  -2(3) + 7

= -6 + 7

= 1

13 and 1 are not equal

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

5 + y - 2 - 6y + y; 8y + 3

<em>Substitute 3 for y in 5 + y - 2 - 6y + y</em>

This gives 5 + 3 - 2 - 6(3) + 3

= 5 + 3 - 2 - 18 + 3

= -9

<em>Then Substitute 3 for y in 8y + 3</em>

This gives  8(3) + 3

= 24 + 3

= 27

-9 and 27 are not equal

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

2y + 7 + y - 2 + y; 4y + 5

<em>Substitute 3 for y in 2y + 7 + y - 2 + y</em>

This gives 2(3) + 7 + 3 - 2 + 3

= 6 + 7 + 3 - 2 + 3

= 17

<em>Then Substitute 3 for y in 4y + 5</em>

This gives  4(3) + 5

= 12 + 5

= 17

17 and 17 are equal

Hence, Option D is correct

6 0
3 years ago
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