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REY [17]
3 years ago
8

Two percent of the customers of a store buy cigars. Half of the customers who buy cigars buy beer. 25 percent who buy beer buy c

igars. Determine the probability that a customer neither buys beer nor buys cigars.
Mathematics
1 answer:
enyata [817]3 years ago
4 0

Answer:

0.95

Step-by-step explanation:

The computation of the probability that a customer neither buys beer nor buys cigars is given below;

Given that, the probabilities are

The customers who purchased cigars be 0.02

The customers who purchased cigars + beer 0.50

And, the customers who purchased beer + cigars be 0.25

Now the probabilities where the customer purchased both

= 0.05 × 0.02

= 0.10

The probability where the customer purchased beer is

= 0.01 ÷ 0.25

= 0.04

Now the probability where a customer neither buys beer nor buys cigars is

= 1 - 0.02 + 0.04 - 0.01

= 0.95

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3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

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k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

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2 years ago
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