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2 years ago

Help answer please!!

Mathematics

1 answer:

12345 [234]2 years ago

6 0

Answer:

h = 19.7 ft.

Step-by-step explanation:

Use the Pythagoras Theorem

$(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2\\\\(20)^2=(3.5)^2+(h)^2\\400=12.25+h^2\\400-12.25=h^2\\387.75=h^2\\$

Now we take square root on both sides,

$\sqrt{387.75} =\sqrt{h^2}\\19.69=h\\h=19.7\ ft\\$

so the value of h equals 19.7 ft.

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If the club decides to send 5 members to the convention, how many different groups of 5 are possible? there are 10 members

hodyreva [135]

For the 1st member, you have 10 possible choices.

For the 2nd, you have 9 remaining choices.

For the 3rd: 8.

So all possible combinations would be:
10*9*8*7*6 = 10! - 5! = 30,240 combinations.

7 0

2 years ago

A=8, b=5, C=90 degree; Find c, A, and B<br><br> This is for the law of sine and cosine

brilliants [131]

Answer:

• c = √89 ≈ 9.434

• A = arcsin(8/√89) ≈ 57.995°

• B = arcsin(5/√89) ≈ 32.005°

Step-by-step explanation:

By the law of cosines, ...

c² = a² + b² -2ab·cos(C)

Since c=90°, cos(C) = 0 and this reduces to the Pythagorean theorem for this right triangle.

c = √(8² +5²) = √89 ≈ 9.434

Then by the law of sines (or the definition of the sine of an angle), ...

sin(A) = a/c·sin(C) = a/c = 8/√89

A = arcsin(8/√89) ≈ 57.995°

sin(B) = b/c·sin(C) = b/c = 5/√89

B = arcsin(5/√89) ≈ 32.005°

6 0

3 years ago

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$

SD = 1.033

4 0

3 years ago

Pls help me guys :( you guys are talented

4vir4ik [10]

D\

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5 0

3 years ago

Plzzzzz help,.............

mafiozo [28]

With what? Maybe I can help

5 0

2 years ago