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3 years ago

15

Kathleen had a bag of candy-coated chocolates. Of the 55 candies in the bag, 20% were red, 30% were orange, 30% were green, and

20% were blue. Of the blue ones, 20% had a damaged candy coating. Approximately how many of the blue ones had a damaged candy coating?
A.
2
B.
11
C.
22
D.
6

2 answers:

pishuonlain [190]3 years ago

5 0

For this case, we find the amount of each colored candy in the bag, having a total of 55.

Reds:

55 -------------> 100%

R ---------------> 20%

Where "R" represents the amount of red candies.

$R = \frac {20 * 55} {100}\\R = 11$

Oranges:

$O = \frac {30 * 55} {100}$

O = 16.5, approximately 16 orange candies.

Green:

$G = \frac {30 * 55} {100}$

G = 16.5, approximately 16 green candies.

Blues:

$B = \frac {20 * 55} {100}\\B = 11$

Now, having a total of 11 blue candies, we must find 20% to know how many have the damaged candy coating.

11 ---------------> 100%

x -----------------> 20%

Where "x" represents the quantity of blue candies with the damaged candy coating

$x = \frac {20 * 11} {100}\\x = 2.2$

So, of 2 blue candies have the damaged candy coating

Answer:

two

Vikentia [17]3 years ago

3 0

Answer:

2

Step-by-step explanation:

55 x 0.2 = 11

11 x 0.2 = 2.2

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

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Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

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Answer:

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Answer:

91.02% probability of selling more than 4 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. Either it is sold, or it is not. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 14, p = 0.5$

Compute the probability of selling more than 4 properties in one week.

Either you sell 4 or less properties in one week, or you sell more. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want to find $P(X > 4)$ . So

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.5)^{0}.(0.5)^{14} = 0.000061$

$P(X = 1) = C_{14,1}.(0.5)^{1}.(0.5)^{13} = 0.000854$

$P(X = 2) = C_{14,2}.(0.5)^{2}.(0.5)^{12} = 0.0056$

$P(X = 3) = C_{14,3}.(0.5)^{3}.(0.5)^{11} = 0.0222$

$P(X = 4) = C_{14,4}.(0.5)^{4}.(0.5)^{10} = 0.0611$

So

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000061 + 0.000854 + 0.0056 + 0.0222 + 0.0611 = 0.0898$

Finally

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.0898 = 0.9102$

91.02% probability of selling more than 4 properties in one week.

8 0

3 years ago