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3 years ago

What is the slope of the line that contains the points (7,-1) and (-2,2)?

Mathematics

1 answer:

Vesnalui [34]3 years ago

8 0

Answer: -1/3

(7,-1)= (x1,y1)

(-2,2)= (x2,y2)

Slope= $\frac{y2-y1}{x2-x1}$

= $\frac{2-(-1)}{-2-7}$

= $\frac{1}{-3}$

1/-3 is the same thing as -1/3

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A large city newspaper periodically reports the mean cost of dinner for two people at restaurants in the city. The newspaper sta

inysia [295]

Answer: C) 133

Step-by-step explanation:

The formula to find the sample size is given by :-

$n=(\dfrac{z^*\cdot\sigma}{E})^2$

, where z* = Critical z-value

$\sigma$ = Population standard deviation for prior study.

E= Margin of error.

As per given , we have

$\sigma=$35$

E= 5

The critical z-value for 90% confidence level is 1.645.

Substitute al;l the value sin the above formula , we get

$n=(\dfrac{1.645\times 35}{5})^2$

$n=(\dfrac{57.575}{5})^2$

$n=(11.515)^2$

$n=132.595225\approx133$

Hence, the minimum sample size needed is 133.

Thus , the correct answer is : C) 133

6 0

3 years ago

Find x and y ! <br><br> pls help giving 50 points !!

Natasha_Volkova [10]

Answer:

y = 20°

x = 35°

Explanation:

Equation's:

1) 2y + x + 105° = 180°

2) 3x + x + 2y = 180°

Make y subject in equation 2

3x + x + 2y = 180

4x + 2y = 180

2y = 180 - 4x

y = 90 - 2x

Insert this into equation 1

2(90 - 2x) + x + 105° = 180°

180 - 4x + x + 105 = 180

-3x = -105

x = 35°

Find value of y

y = 90 - 2x

y = 90 - 2(35)

y = 20°

4 0

1 year ago

Read 2 more answers

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$