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3 years ago

150 students were surveyed to see what sport they play what percentage plays soccer

1 answer:

7 0

Not enough info.
You need to know how many played soccer.

If the number who played soccer is represented by S;

Then percentage = S/150 x 100

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If ~ p: 2 + 2 =4

Art [367]

Answer:

A. $\displaystyle 2 \times 2 = 4$

Step-by-step explanation:

I could not make clarifications of what you are asking, but this is similar to $\displaystyle 2 + 2 = 4.$

I am joyous to assist you anytime.

6 0

3 years ago

Read 2 more answers

(Refer to the attached image)

gayaneshka [121]

Explanation:

Provided <u>2 length sides</u> and <u>one angle</u> also need to find <u>one missing side</u>.

So, use cosine rule:

a² = b² + c² - 2bc cos(A)

c² = 9² + 11² - 2(11)(9) cos(57)

c² = 94.16147

c = √94.16147 = 9.70 cm

d² = 5² + 7² - 2(5)(7) cos(48)

d² = 27.16

d = √27.16 = 5.21

5² = 7² + 9² - 2(7)(9) cos(H)

-126cos(H) = 25 - 49 - 81

cos(H) = -105/-126

cos(H) = 5/6

H = cos⁻¹(5/6) = 33.56°

8² = 4² + 7² - 2(4)(7) cos(J)

-56cos(J) = 64 - 16 - 49

cos(J) = -1/-56

J = cos⁻¹(1/56) = 88.98°

Passcode: 3142

8 0

2 years ago

Ask Your Teacher Consider a binomial random variable with n = 5 and p = 0.8. Let x be the number of successes in the sample. Eva

Readme [11.4K]

Answer:

0.942 is the required probability.

Step-by-step explanation:

We are given the following in the question:

x is a binomial random variable with n = 5 and p = 0.8.

Then,

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

$P(x \geq 3) = P(x = 3) + P(x = 4) +P(x=5)\\\\= \binom{5}{3}(0.8)^3(1-0.8)^2 +\binom{5}{4}(0.8)^4(1-0.8)^1 +\binom{5}{5}(0.8)^5(1-0.8)^0\\\\= 0.2048 +0.4096+0.3276=0.942$

0.942 is the required probability.

4 0

4 years ago

Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

WITCHER [35]

Answer:

Required solution (a) $f(x,y,z)=xyz+3z^2+C$ (b) 40.

Step-by-step explanation:

Given,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k$

(a) Let,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}$

Then,

$f_x=yz,f_y=xz,f_z=xy+6z$

Integrating $f_x$ we get,

$f(x,y,z)=xyz+g(y,z)$

Differentiate this with respect to y we get,

$f_y=xz+g'(y,z)$

compairinfg with $f_y=xz$ of the given function we get,

$g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)$

Then,

$f(x,y,z)=xyz+h(z)$

Again differentiate with respect to z we get,

$f_z=xy+h'(z)=xy+6z$

on compairing we get,

$h'(z)=6z\implies h(z)=3z^2+C$ (By integrating h'(z)) where C is integration constant. Hence,

$f(x,y,z)=xyz+3z^2+C$

(b) Next, to find the itegration,

$\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40$

3 0

3 years ago

What is greater, 2/5, 3/10, 1/2, or 1/10?

Pani-rosa [81]

2/5 = 40%
3/10= 30%
1/2= 50%
1/10= 10%

so it's 1/2

7 0

3 years ago

Read 2 more answers

150 students were surveyed to see what sport they play what percentage plays soccer￼￼

150 students were surveyed to see what sport they play what percentage plays soccer