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3 years ago

11

Find the sine of ∠A.

2 answers:

alisha [4.7K]3 years ago

7 0

Answer: The value of sin∠A = 36.86°.

Step-by-step explanation:

Since we have given that

AB = 5 units

AC = 4 units

BC = 3 units

So, we need to find the sine ∠A.

As we know the formula for "Sine":

$\sin A=\dfrac{Opposite}{Adjacent}\\\\\sin A=\dfrac{BC}{AB}\\\\\sin A=\dfrac{3}{5}\\\\\sin A=0.6\\\\A=\sin^{-1}((0.6)\\\\A=36.86^\circ$

Hence, sin∠A = 36.86°.

drek231 [11]3 years ago

6 0

Sin A=(opposite leg)/hypotenuse= 3/5=0.6
For angle A , opposite leg is BC=3
Hypotenuse = 5

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Im trying to find the measure of the numbered angle:)??

bazaltina [42]

3 and 2 would be 122 since 3 and the 58° are a linear pair making them add up to 180. And 3 and 2 are vertical so they would be congruent.
4 would then have to be 38 since the three angles would have to measure up to 180.
And the 1 would have to be 36 with the same reasoning as before.
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4 0

3 years ago

Read 2 more answers

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.

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3 years ago

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3 years ago

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xxTIMURxx [149]

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alekssr [168]

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2 years ago