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3 years ago

Find value of x such that the data set has a mean of 105: 99, 119, 106, 112; 108

1 answer:

6 0

I'm assuming that x is part of the data set, and, with x, the mean equals 105. To find the value of x, you must add all the data values together to get 544+x (you still don't know what x equals). Then put 544+x over how many days values there are, including x (there are 6). You should have 544+x/6. Now, as this is how you would calculate the mean if you knew what x was equal to, you must set it equal to the mean, since you know what it is (105). You should now have 544+x/6 = 105. You have your equation set up--now you just have to solve it. I would multiply by 6 on both sides to get rid of the 6 on the left side. You would then have 544+x = 630. I would finally subtract 544 from both sides to get x = 86. Your final answer is x = 86.

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Select the property of equality used to arrive at the conclusion. if 1/3 x=5 then x=15 the multiplication property of equality t

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Depends on what number(s)/fraction(s) your working with.

Step-by-step explanation:

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3 years ago

part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.

Nat2105 [25]

Answer: $\bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}$

Step-by-step explanation:

Pythagorean Theorem is: a² + b² = c² , where "c" is the hypotenuse

$9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}$

Note: 4² + (8√2)² = hypotenuse² → hypotenuse = 12

$10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}$

Note: 12² + opposite² = 20² → opposite = 16

$11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: adjacent² + 5² = 6² → adjacent = √11

$12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}$

Note: adjacent² + 7² = (13√2)² → adjacent = 17

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Step-by-step explanation:

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