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Thepotemich [5.8K]
3 years ago
15

The cost of vacation to a cabin resort for a night is $95 for each person. Each cabin also has to pay a $25 recreational equipme

nt rental fee. Model the cost, C, for staying x nights at the resort.
Mathematics
2 answers:
Rom4ik [11]3 years ago
8 0

x=95x+ 25

Step-by-step explanation:

Blababa [14]3 years ago
7 0
(x) = 95x + 25 is correct, since the $25 cost is fixed, but the $95 cost increases as x increases.
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Solve the following equations x1/3=2 <br> Y-2=9<br> (2s)1/2=9<br> 2n-1=16
Nookie1986 [14]

x \times  \frac{1}{3}  = 2 \\  =  > x = 2 \times 3 \\  =  > x = 6

y - 2 = 9 \\  =  > y = 9 + 2 = 11

(2s) \frac{1}{2}  = 9 \\  =  > s = 9

2n - 1 = 16 \\  =  > 2n = 16 + 1 \\  =  > 2n = 17 \\  =  > n =  \frac{17}{2}  = 8.5

<h3>The answers are :</h3><h3>x = 6</h3><h3>y = 11</h3><h3>s = 9</h3><h3>n = 8.5</h3><h3>Hope it helps!</h3><h3 />

7 0
2 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
ST | PR. Find QR. 35 28 40
Greeley [361]

Answer:

The answer is 47

Step-by-step explanation:

I'm pretty sure this is the answer :)

40 - 28 = 12

35 + 12 = 47

Proof: 35 - 28 = 7

          47 - 40 = 7

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Solve equation -3m=4m - 15
timama [110]

Answer:

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Step-by-step explanation:

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