MATH QUESTION ON SLOPESSSSSSS

Mathematics

1 answer:

alex41 [277]3 years ago

3 0

I’m fairly sure it is the last one. But the pictures are hard to see. 3/2 is the slope and -1 is the y intercept. Find the point at (0,-1) , the y intercept. Then go up 3 and over 2 to find the next point

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Graph triangle cuz with vertices x(-2, -4), y(2, -1), and z(4, -3) on the coordinate grid.

Vera_Pavlovna [14]

This is what I got pls mark me brainlest

4 0

2 years ago

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Maurinko [17]

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .

6 0

3 years ago

A gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than

Neko [114]

Answer:

The probability that no more than 70% would prefer to start their own business is 0.1423.

Step-by-step explanation:

We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.

Let $\hat p$ = sample proportion of people who prefer to start their own business

The z-score probability distribution for the sample proportion is given by;

Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, p = population proportion who would prefer to start their own business = 72%

n = sample of 18-29 year-olds = 600

Now, the probability that no more than 70% would prefer to start their own business is given by = P( $\hat p$ $\leq$ 70%)

P( $\hat p$ $\leq$ 70%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } }$ ) = P(Z $\leq$ -1.07) = 1 - P(Z < 1.07)