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2 years ago

6

I need help right away ASAP!!!Please help me please finding the missing angle

2 answers:

german2 years ago

8 0

$\\ \sf\longmapsto tan\Theta=\dfrac{P}{B}$

$\\ \sf\longmapsto tan40=\dfrac{48}{x}$

$\\ \sf\longmapsto 0.8=\dfrac{48}{x}$

$\\ \sf\longmapsto 0.8x=48$

$\\ \sf\longmapsto x=\dfrac{48}{0.8}$

$\\ \sf\longmapsto x=60$

Inessa [10]2 years ago

6 0

Answer:

x = 57.2

Step-by-step explanation:

first : the third angle is (the sum of all angles in a triangle is always 180 degrees) : 180 - 90 - 40 = 50 degrees.

law of sines :

x/sin(50) = 48/sin(40)

x = 48 × (sin(50)/sin(40)) = 57.2

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3 years ago

NEED HELP ASAP GIVING BRAINLIST!!!!!!!!

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3 years ago

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Expand and simplify (2x^2)(4x+5)-8x(x^2-2)

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5 0

2 years ago

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

1. The company has agreed on a 9- character password format, comprised of uppercase letters, lower case letters, and digits (0-9

pochemuha

Answer: a) 1977060966, b) 146581344.

Step-by-step explanation:

Since we have given that

There are 10 numbers,

There are 26 lower case letters,

There are 26 upper case letters,

Since the first 3 characters have to be digits and next 6 can be upper or lower.

So, a) repetitions are allowed:

$10^3\times (26+26)^6\\\\=10^3\times 52^6\\\\=1977060966$

b) repetitions are not allowed:

$^{10}P_3\times ^{52}C_6\\\\=720\times 20358520\\\\=146581344$

Hence, a) 1977060966, b) 146581344.

5 0

3 years ago