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Viefleur [7K]
4 years ago
11

A cab charges $1.50 for the 1st mile and $.50 for each additional mile. Right and solve an inequality to determine how many mile

s Sharon can travel if she is $25 to spend
Mathematics
1 answer:
vekshin14 years ago
4 0
Inequality
25 \geqslant 1.5 + .5x
next-
subtract 1.5 from each side
23.5 \geqslant .5x
next-
divide .5 from each side to get answer
47 \geqslant x
the answer is 47 miles




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What are the first three terms of the sequence defined by the recursive function an = an-1-(an-2-4), when a5 =-2 and a6 = 0?
Wittaler [7]

Answer:

<em>6, 10 and 8</em>

Step-by-step explanation:

Given the recursive function an = an-1-(an-2 - 4), when a5 =-2 and a6 = 0?

a6 = a5 - (a4 - 4)

0 = -2  - (14 - 4)

2 = - (a4 - 4)

-2 = a4 - 4

a4 = -2 + 4

a4 = 2

a5 = a4 - (a3 - 4)

-2 = 2  - (a3 - 4)

-2-2 = - (a3 - 4)

-4 = -(a3 - 4)

4 = a3 - 4

a3 = 4+4

a3 = 8

Also

a4 = a3 - (a2 - 4)

2 = 8 - (a2 - 4)

2-8 = - (a2 - 4)

-6 = -(a2 - 4)

6 = a2 - 4

a2 = 6+4

a2 = 10

a3 = a2 - (a1 - 4)

8 = 10 - (a1 - 4)

8-10 = - (a1 - 4)

-2 = -(a1 - 4)

2 = a1 - 4

a1 = 2+4

a1 = 6

<em>Hence the first 3 terms are 6, 10 and 8</em>

6 0
3 years ago
Read 2 more answers
The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for
Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

<em>So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;</em>

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s^{2} = sample variance = 1.88

           \sigma^{2} = population variance

            n = sample of windshields = 83

So, 95% confidence interval for population variance, \sigma^{2} is;

P(58.85 < \chi^{2} __8_2 < 108.9) = 0.95 {As the table of \chi^{2} at 82 degree of freedom

                                              gives critical values of 58.85 & 108.9}

P(58.85 < \frac{(n-1)s^{2} }{\sigma^{2} } < 108.9) = 0.95

P( \frac{ 58.85}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{108.9}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{108.9 } < \sigma^{2} < \frac{ (n-1)s^{2}}{58.85 } ) = 0.95

<em><u>95% confidence interval for</u></em> \sigma^{2} = ( \frac{ (n-1)s^{2}}{108.9 } , \frac{ (n-1)s^{2}}{58.85 } )

                                                  = ( \frac{ (83-1)\times 1.88}{108.9 } , \frac{ (83-1)\times 1.88}{58.85 } )

                                                  = (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

8 0
3 years ago
What equation has a slope of 2 and contains (1,4)
user100 [1]

Step-by-step explanation:

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What is 15 5/7+3/7?<br> i need help!
kobusy [5.1K]

Answer:

16 1/7

Step-by-step explanation:

simple add

start with the fractions

5/7+3/7 = 8/7

simplify 1 1/7

add your whole numbers 15+1= 16 and the add your fraction 16+1/7= 16 1/7

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A spherical water tank has a diameter of 10 feet. How much water does it hold when half full? (Use 3.14 for p). . A. 261.66 cubi
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If a spherical water tank has a diameter of 10 feet, it will hold approximately 261.66 cubic feet when it is half full. The correct answer is A. 
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