Suppose that from a group of 30 athletes, 17 compete in a 100-meter race, 8 compete in a

Solve to find the answer of blue pills Mr body took: number of blue pills, b, mr mattics give him was 3 more than the number of

barxatty [35]

Answer:

Mr body took 9 blue pills

Step-by-step explanation:

Total number of pills consumed = 17

Number of blue pills, b = 11

Number of blue pills, b = g + 3

So that,

b = g + 3

11 = g + 3

g = 8

Total number of pills recommended = b + g

= 11 + 8

= 19

Total number of pills consumed = 17

Thus,

Number of blue pills consumed = 17 - 8

= 9

Therefore, Mr body took 9 blue pills.

5 0

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

3 years ago

What is 10/8 divided by 1/4

daser333 [38]

5

Mark brainliest please

Hope this helps you

6 0

2 years ago

Read 2 more answers

2 people can plough a field in 6 hours.

Tems11 [23]

B is the answer because it’s the one that makes most sense

5 0

3 years ago

Read 2 more answers

Three friends each have some ribbon. Carol has 54 inches of ribbon, Tino has 7.5 feet of ribbon, and Baxter has 3.5 yards of rib

Nastasia [14]

1 ft = 12 inches......1 yd = 36 inches

carol has 54 inches
Tino has 7.5 ft......7.5 * 12 = 90 inches
baxter has 3.5 yds......3.5 * 36 = 126 inches
for a total of : (54 + 90 + 126) = 270 inches total <===

1 ft = 12 inches....so 270 inches = 270/12 = 22.5 ft <==
1 yd = 36 inches..so 270 inches = 270/36 = 7.5 yds <==

5 0

3 years ago