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trapecia [35]
3 years ago
8

Josh and Anthony have a lemonade stand.They charge $1 for 2 cups of lemonade.They sell 14 cups each afternoon.Is it reasonable t

o say Josh and Anthony earned more than $50 after 3 afternoons?
Mathematics
1 answer:
Rudik [331]3 years ago
6 0
No it is not, whatsoever.

$1 for 2 cups. 14 cups in an afternoon means they got $7. If they did this for three days (7 x 3) they'd have 21 dollars. it is not reasonable because 21 is nowhere near 50.

Hope this helps

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Answer: 4 - (xxx/555)

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At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more
I am Lyosha [343]

Answer:

The probability is  P( X \le 4 ) = 0.0054

Step-by-step explanation:

From the question we are told that

   The percentage that are on time is  p =  0.76

   The  sample size is n =  11

   

Generally the percentage that are not on time is

     q =  1- p

     q =  1-  0.76

     q = 0.24

The  probability that no more than 4 of them were on time is mathematically represented as

        P( X \le 4 ) =  P(1 ) +  P(2) + P(3) +  P(4)

=>     P( X \le 4 ) =  \left n } \atop {}} \right.C_1 p^{1}  q^{n- 1} +   \left n } \atop {}} \right.C_2p^{2}  q^{n- 2} +  \left n } \atop {}} \right.C_3 p^{3}  q^{n- 3}  +  \left n } \atop {}} \right.C_4 p^{4}  q^{n- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{11- 1} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{11- 2} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{11- 3}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{11- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{10} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{9} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{8}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{7}

= \frac{11! }{ 10! 1!}  (0.76)^{1}  (0.24)^{10} +   \frac{11!}{9! 2!}  (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!}  (0.76)^{3}  (0.24)^{8}  + \frac{11!}{7!4!}  (0.76)^{4}  (0.24)^{7}

P( X \le 4 ) = 0.0054

4 0
3 years ago
Find the variance of the set of data to the nearest tenth: 5, 8, 2, 9, 4
Orlov [11]

Answer:

Therefore the variance on the data set is 8.3

Step-by-step explanation:

In order to find the variance of the set of data we first need to calculate the mean of the set, which is given by:

mean = sum of each element / number of elements

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We can now find the variance by applying the following formula:

s^{2} = \frac{\sum_{(i=1)}^n(x_i - mean)^2}{n -1}

So applying the data from the problem we have:

s² = [(5 - 5.6)² + (8 - 5.6)² + (2 - 5.6)² + (9 - 5.6)² + (4 - 5.6)²]/(5 - 1)

s² = [(-0.6)² + (2.4)² + (-3.6)² + (3.4)² + (-1.6)²]/4

s² = [0.36 + 5.76 + 12.96 + 11.56 + 2.56]/4 = 8.3

Therefore the variance on the data set is 8.3

7 0
2 years ago
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