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3 years ago

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The table shows the snack preferences of 50 shoppers at the mall. what is the probability that one shopper, selected at random f

rom the 50 surveyed, preferred the potato chips or pretzels?
A.1/5
B.13/25
C.1/10
D.18/25

2 answers:

Ipatiy [6.2K]3 years ago

6 0

1/5 is the answer homie

Gala2k [10]3 years ago

3 0

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Number of people who prefer chips = 12

Number of people who prefer pretzels = 14

Total number of people = 50

So, Probability that one preferred the potato chips or pretzels is given by

$\dfrac{12}{50}+\dfrac{14}{50}\\\\=\dfrac{26}{50}\\\\=\dfrac{13}{25}$

Probability that one preferred the potato chips or pretzels is $\dfrac{13}{25}$

Hence, Option 'B' is correct.

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Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

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To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

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Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

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$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

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$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

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