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vitfil [10]
2 years ago
5

In a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979,

two groups of childless wives aged 25 to 29 were selected at random, and each was asked if she eventually planned to have a child. One group was selected from among wives married less than two years and the other from among wives married five years. Suppose that 240 of the 300 wives married less than two years planned to have children some day compared to 288 of the 400 wives married five years. Can we conclude that the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years? Make use of a P -value.
Mathematics
1 answer:
Fed [463]2 years ago
4 0

Answer:

we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years

Step-by-step explanation:

Given that in a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups of childless wives aged 25 to 29 were selected at random, and each was asked if she eventually planned to have a child. One group was selected from among wives married less than two years and the other from among wives married five years.

Let X be the group married less than 2 years and Y less than 5 years

                         X        Y     Total

Sample size   300   300    600

Favouring       240   288    528

p                      0.8     0.96  0.88

H_0: p_x=p_y\\H_a: p_x>p_y

p difference = -0.16

Std error for difference = \sqrt{0.88*0.12/600} =0.01327

Test statistic = p difference/std error=-6.03

p value <0.000001

Since p is less than alpha 0.05 we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years

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Allison plays basketball and last season she made a total of 87 shots (two- or three- point shots). Her points total was 191. Ho
yarga [219]

Answer:

17

Step-by-step explanation:

Let the number of 2 point shots be t.

Let the number of 3 point shots be h.

Allison made 87 shots and scored 191 total points. This implies two things:

t + h = 87 ________(1)

2t + 3h = 191 ______(2)

Let us eliminate t by multiplying (1) by 2 and subtracting from (2):

=> 2t + 3h = 191

 -  <u>2t + 2h = 174</u>

     <u>          h  = 17 </u>

<u />

Therefore, she made 17 three point shots.

4 0
2 years ago
The vertex of this parabola is at (-2,-3). When the x-value is -1, the y-value is -5. What is the coefficient of the squared exp
elena55 [62]

Answer:

Option C is correct

Step-by-step explanation:

Given: vertex of this parabola is at (-2,-3)

To find: coefficient of the squared expression in the parabola’s equation if the x-value is -1, the y-value is -5

Solution:

The equation of parabola is of the form y=a(x-h)^2+k

Here, a is the coefficient of the squared expression in the parabola’s equation.

Put (h,k)=(-2,-3)\,,\,(x,y)=(-1,-5)

-5=a(-1+2)^2-3\\-5+3=a(1)^2\\-2=a\\a=-2

So, the coefficient of the squared expression in the parabola’s equation is -2

6 0
2 years ago
Triangle cde maps to triangle lmn with the transformation (x,y) —&gt; (x+3,y-2) —&gt; (2/3x, 2/3y)
Mamont248 [21]
The answer is 890 best answer
4 0
2 years ago
Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
Ksju [112]
Well, here's one way to do it at least... 

<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>

<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>

<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>

<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>

<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>

<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>

<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>

<span>Now apply the Law of Sines to find the other two angles. </span>

<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>

<span>Two down, one to go... </span>

<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>

<span>So your three angles are: </span>

<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
8 0
2 years ago
Leticia charges $8 per hour to babysit. She babysat Friday night for 4 hours,
JulsSmile [24]

For this case we have that the variable "x" represents the number of hours that Leticia uses to take care of children on Saturday.

IF on Friday I use 4 hours ($ 8 each) and on Saturday "x" hours ($ 8 each) obtaining a profit of $ 72, we have the following equation:

8 (4 + x) = 72

We apply distributive property:

32 + 8x = 72\\8x = 72-32\\8x = 40\\x = \frac {40} {8}\\x = 5

So, on Saturday she spent 5 hours.

Answer:

8 (4 + x) = 72\\x = 5

6 0
2 years ago
Read 2 more answers
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