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3 years ago

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At a party, there are 2 six-packs of regular cola, 1 six-pack of diet cola, 1 six-pack of cherry cola, and 1 six-pack of vanilla

cola. If a can of cola is chosen at random, what is the probability it will be a cherry cola or a vanilla cola?
A. 1/5 B. 2/5 C. 1/4 D. 1/2 Please select the best answer from the choices provided

A B C D

1 answer:

aliya0001 [1]3 years ago

8 0

Answer:

B.

Step-by-step explanation:

There are 5 total 6-packs of cola. 1 is cherry and the other is vanilla. 2 out of 5 are the flavors. This means you have a 2 in 5 chance of getting a cherry or vanilla cola.

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A parallelogram has an area of 100 square units. Its perimeter is

Nonamiya [84]

good morning ☕️

___________________

Step-by-step explanation:

—Dimension1 =25 and dimension2 =4

Area=25*4=100

Perimeter=2(25+4)=2(29)=58.

—-Dimension1 =12.5 and dimension2 =8

Area=12.5*8=100

Perimeter=2(12.5 + 8)=2(20.5)=41

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4 years ago

Which of the following describes the square root of 41. 5,6 6,7 20,21 40,42

cestrela7 [59]

Answer:

6,7

Step-by-step explanation:

the squre root of 41 is 6.403

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3 years ago

How to do exponents with variable

storchak [24]

Answer:

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Step-by-step explanation:

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3 years ago

Given the sequence 10,20,40

Marta_Voda [28]

50,70,80,100,110,130 that would be it

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3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

4 years ago