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3 years ago

Find the distance between the pair of points.. (-6, -5), (7, -8)

Mathematics

1 answer:

horrorfan [7]3 years ago

6 0

Answer:

$\sqrt{178}$ units

Step-by-step explanation:

Use the distance formula $d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$ where $d$ is the positive distance between $(x_1,y_1)$ and $(x_2,y_2)$ :

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

$d=\sqrt{(-8-(-5))^2+(7-(-6))^2}$

$d=\sqrt{(-8+5)^2+(7+6)^2}$

$d=\sqrt{(-3)^2+(13)^2}$

$d=\sqrt{9+169}$

$d=\sqrt{178}$

Therefore, the distance between $(-6,-5)$ and $(7,-8)$ is $\sqrt{178}$ units.

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3 years ago

Evaluate the expression. 33 + 6( 2 + 36 )

Alinara [238K]

Answer: 261

Step-by-step explanation:

You use PEMDAS to solve this problem

you first do the Parenthesis (p)

33+6(2+36)

33+6 (38)

We then multiply first (M)

33 + 6(38) - we know that 8 times 6 is 48 ; 6 times 3 is 18 +4 because we brought down the 4 from 48 to 3. which will be equal to 228

33 + 228

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3 years ago

If the x-intercepts are (-5,0) and (3.0), then the x-coordinate of the vertex is

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If this is for a parabola, then the x-coordinate is -1, since it is directly in the middle of -5 and 3, with 4 spaces on each side (symmetry). The equation is just x = -1 since it just shows the equation of the vertical line which splits the parabola into two equal parts

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3 years ago

Find the area of the rectangle, to the nearest hundredth, with a base of 5 inches and a height = 2.5 inches.

Juli2301 [7.4K]

The equation for the area of a rectangle is: A = bh, where A = area, b = length of base, and h = height.

You're told that the base, b = 5 in, and the height, h = 2.5 in. Plug these numbers into the equation to find the area:
A = bh
A = (5)(2.5)
A = 12.5 $in^2$

If you include the hundredth place, your answer would be 12.50 $in^2$ .

-------

Answer: 12.50 $in^2$

5 0

3 years ago

The Highway Safety Department wants to study the driving habits of individuals. A sample of 37 cars traveling on a particular st

kodGreya [7K]

Answer:

90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

Step-by-step explanation:

We are given that a sample of 37 cars traveling on a particular stretch of highway revealed an average speed of 70.7 miles per hour with a standard deviation of 6.3 miles per hour.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average speed of cars = 70.7 miles per hour

s = sample standard deviation = 6.3 miles per hour

n = sample of cars = 37

$\mu$ = true mean speed

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 90% confidence interval for the true mean, $\mu$ is ;

P(-1.688 < $t_3_6$ < 1.688) = 0.90 {As the critical value of t at 36 degree of

freedom are -1.688 & 1.688 with P = 5%}

P(-1.688 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.688) = 0.90

P( $-1.688 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.688 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.688 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.688 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.688 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.688 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $70.7-1.688 \times {\frac{6.3}{\sqrt{37} } }$ , $70.7+1.688 \times {\frac{6.3}{\sqrt{37} } }$ ]

= [68.9517 miles per hour , 72.4483 miles per hour]

Therefore, 90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

The interpretation of the above interval is that we are 90% confident that the true mean speed of all cars will lie between 68.9517 miles per hour and 72.4483 miles per hour.

8 0

3 years ago