Evaluate f(x)=-7x+3 when x=-2, x=0, and x=5

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

The cost of renting a chain saw is $3.90 per hour plus $6.50 for a cán

puteri [66]

35.75
explanation-
3.90 ph x 7.5 hours = 29. 25
plus the gas can 6.50
29.25+6.50= 35.75

5 0

3 years ago

Please help meeeee!!

Xelga [282]

Answer:

-27

Step-by-step explanation:

order of operations 2÷2 = 1 +53 =54

3x1 = 3 -5 = -2

54÷-2 = -27

5 0

3 years ago

Denise is designing the seating arrangement for a concert an outdoor theater. To give everyone a good view, each row must have 6

Soloha48 [4]

Given:
1st term = 11
common difference = 6

f(x) = 11 + 6(x - 1)
f(18) = 11 + 6(18-1)
f(18) = 11 + 6(17)
f(18) = 11 + 102
f(18) = 113 number of seats in row 18.

Row
 1 11 11
2 11 6 17
 3 17 6 23
 4 23 6 29
 5 29 6 35
 6 35 6 41
 7 41 6 47
 8 47 6 53
 9 53 6 59
 10 59 6 65
 11 65 6 71
 12 71 6 77
 13 77 6 83
 14 83 6 89
 15 89 6 95
 16 95 6 101
 17 101 6 107
 18 107 6 113

6 0

4 years ago

Anyone! Please help me! I don't understand at all what to do! Help would be really appreciated :)

Rom4ik [11]

Answer: They are parallel

Step-by-step explanation:

If two lines are parallel , then they must have the same slope and if two lines are perpendicular , the product of their slope must be -1.

To check this , we must calculate the slope of the two lines given.

Slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$