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Ahat [919]
3 years ago
12

Solve the following equation for x: z=x^2+wx

Mathematics
1 answer:
Alja [10]3 years ago
4 0
Minus z both sides
0=x^2+wx+z
we can use quadratic formula or manually complete the square
let's use quadratic

for
0=ax^2+bx+c
x= \frac{-b+/- \sqrt{b^2-4ac} }{2a}
given
0=1x^2+wx-z
a=1
b=w
c=-z

x= \frac{-w+/- \sqrt{w^2-4(1)(-z)} }{2(1)}
x= \frac{-w+/- \sqrt{w^2+4z} }{2}

x= \frac{-w+ \sqrt{w^2+4z} }{2} or x= \frac{-w- \sqrt{w^2+4z} }{2}
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The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization’s staff of
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Null Hypothesis, H_0 : p \geq 45%    

Alternate Hypothesis, H_A : p < 45%

Step-by-step explanation:

We are given that the proportion of women in similar sales positions across the country in 2004 is less than 45%.

The collected random sample of size 50 showed that only 18 were women.

<u><em>Let p = proportion of women in similar sales positions across the country in 2004</em></u>

So, Null Hypothesis, H_0 : p \geq 45%    

Alternate Hypothesis, H_A : p < 45%

Here, <u><em>null hypothesis states that</em></u> the proportion of women in similar sales positions across the country in 2004 is more than or equal to 45%.

On the other hand, <u><em>alternate hypothesis states that</em></u> the proportion of women in similar sales positions across the country in 2004 is less than 45%.

The test statistics that would be used here is One-sample z proportion test statistics, i.e;

                        T.S.  =  \frac{\hat p -p}{\sqrt{\frac{\hta p(1-\hat p)}{n} } }  ~ N(0,1)

Hence, the above hypothesis is appropriate for the given situation.

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3 years ago
A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form
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Answer:

\vec{v}= \text{ or } \approx

Step-by-step explanation:

Component form of a vector is given by \vec{v}=, where i represents change in x-value and j represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector \vec{v}=, the magnitude is ||v||=\sqrt{i^2+j^2.

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being i, one leg being j, and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}

To find the other leg, i, we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}

Verify that (9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark

Therefore, the component form of this vector is \vec{v}=\boxed{}\approx \boxed{}

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