The following data show the prices of different types of outfits at a store: $2, $2, $28, $26, $25, $27, $25, $27, $26, $28, $30
Tcecarenko [31]
2, 2, 25, 25, 26, 26, 27, 28, 28, 30
Minimum: 2
Maximum: 30
Median: 26
Lower quartile: 25
<span>Upper quartile: 28</span><span>
The box plot will have its left tail longer than the right tail because a few exceptionally low prices make the distribution skewed to the left.</span>
49. From 3 coin tosses, there are 8 possible outcomes:
... TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
All but the first have at least one head, so 7/8 of the possibilities have at least one head. That's 87.5% (not among your choices).
Likewise, all but the last listed outcome have at least one tail. The problem is symmetrical that way when the coin is fair. 87.5% of outcomes have at least one tail.
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Perhaps you can tell I read your question as having two parts. If your question is the probability of getting at least one head AND at least one tail, you can see that condition includes 6 of the 8 outcomes, or 75%, matching selection d.
50. See for yourself: the calculator says 66.82%. Your best choice is selection d.
Answer:
Description
Step-by-step explanation:
1. Factor 2x^2+11x+12 -> (2x+3)(x+4)
x=-4, -3/2
Both numbers are rational numbers as both can be expressed in the form in a quotient with integer values.
2. If discriminant is negative, then solutions are imaginary, meaning that the discriminant is less than 0, so you can set up this equation.
11^2 - 4(a)(12) < 0
121 - 48a < 0
a>121/48
The closest integer to 121/48 (~2.52) such that it satisfies the above conditions is 3, so the lowest value for a is 3.
Change the wording a bit, and type in your answer.
60 first 20 second 15 third 10 fourth nothing fifth