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irina [24]
2 years ago
12

Jenny wants to rent a truck for one day. She contacted two companies. Laguana’s Truck Rentals charges $20 plus $2 per mile. Salv

atori’s truck rentals charges $3 per mile. After how many miles will the total cost for both companies be the same?
Mathematics
1 answer:
zlopas [31]2 years ago
3 0

Answer:

The total cost will be the same for both companies after 20 miles.

Step-by-step explanation:

First, I wrote the equations:

Laguna's Truck Rentals: 2x + 20

Salvatori's truck rentals: 3x

To find when the total cost for both companies will be the same, I kept substituting numbers for the variable until it worked. So, the total cost for both companies will be the same after 20 miles. Hope this helped!

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g In a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained
Marat540 [252]

Answer:

95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

Step-by-step explanation:

We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

                          P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of residents 65-years or older who had obtained a flu shot = \frac{28}{111} = 0.25

          n = sample of residents 65-years or older = 111

          p = population proportion of residents who were getting the flu shot

<em>Here for constructing 95% confidence interval we have used One-sample z test for proportions.</em>

<u>So, 95% confidence interval for the population proportion, p 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-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } },\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

   = [ 0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } } , 0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } } ]

   = [0.169 , 0.331]

Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

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