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Law Incorporation [45]
2 years ago
7

What does 2miles=feet?

Mathematics
2 answers:
Leya [2.2K]2 years ago
8 0
I think 2mile=24 feet
iVinArrow [24]2 years ago
6 0
2 miles is 10560 feet.
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What is the equation of a line, in general form, that passes through point (1, -2) and has a slope of 1/3. 1.x - 3 y - 7 = 2.x -
Nutka1998 [239]

Answer:

x- 3y - 7=0

Step-by-step explanation:

the line in the slope-intercept form:

  • y= mx+b

m= 1/3 and point given (1, -2) by using which we find y-intercept:

  • b= y-1/3x
  • b= -2- 1/3*1= - 7/3
  • y= 1/3x- 7/3

General form of a line is: ax+by+c= 0

converting:

  • y= 1/3x- 7/3  ⇒  3y= x- 7  ⇒  x- 3y - 7=0
5 0
3 years ago
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

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

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

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

<u>95% confidence interval for</u> (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
Undo the half of what was left amount. Multiply 2
34kurt

Answer:

the three numbers below are your answers. (in order)

Step-by-step explanation:

2*2=4\\4+5=9\\9+8=17

6 0
2 years ago
Please help me!
umka2103 [35]

Answer:

C(m) = 0.10x + 41

Step-by-step explanation:

C(m) = 0.10m + 41

7 0
3 years ago
What is the answer to 39=3m-15
Vladimir [108]
39=3m-15
39+15=3m
54=3m
54/3=3m/3
m=18
4 0
2 years ago
Read 2 more answers
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