In the question "Marta’s mother shared a funny video of their cat on the Internet. The total number of people who had viewed the video over the first 30 days could be modeled using the function f(x) = 33(1.3)x where x is the number of days the video has been online. About how many people had viewed the video once it had been online for 20 days?" The correct answer is 6271 because given the function f(x) = 33(1.3)^x where x is the number of days the video has been online. For x = 20 days: f(20) = 33(1.3)^20 = 33(190.0496377) = <span>6,271</span>

3 0

4 years ago

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Mass: 21 g volume:7cm3 density.

AfilCa [17]

D = M/V

D= 21/7

D = 3 g/cm3

3 0

3 years ago

A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not

castortr0y [4]

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

<em><u>Solution:</u></em>

<em><u>The probability is given as:</u></em>

$Probability = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

$P(B) = \frac{8}{12}$

$P(B n R) = P(B) \times P(R)\\\\P(B n R) = \frac{8}{12} \times \frac{4}{11}\\\\P(B n R) = \frac{8}{33}$

<h3><u>Find </u><u> P(R | B)</u></h3><h3> $P(R | B) = \frac{P(R n B)}{P(B)}\\\\P(R | B) = \frac{\frac{8}{33}}{\frac{8}{12}}\\\\P(R | B) = \frac{8}{33} \times \frac{12}{8}\\\\P(R | B) = \frac{4}{11}$ </h3>

<em><u>The probability that the first ball chosen is black and the second ball chosen is red is about percent</u></em>

$\frac{8}{33} \times 100 = 0.24 \times 100 = 24 \%$

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

4 0

4 years ago