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3 years ago

Find the distance between (-8,-2) and (6,-1)

Mathematics

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{ \sqrt{197} \: \: units}}}}}$

Step-by-step explanation:

Let the points be A and B

Let A ( -8 , -2 ) be ( x₁ , y₁ ) and B ( 6 , - 1 ) be ( x₂ , y₂ )

<u>Finding the distance between these points </u>

$\boxed{ \sf{distance = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }}$

$\dashrightarrow{ \sf{ \sqrt{ {(6 - ( - 8))}^{2} + {( - 1 - ( - 2))}^{2} } }}$

$\dashrightarrow{ \sf{ \sqrt{ {(6 + 8)}^{2} + {( - 1 + 2)}^{2} } }}$

$\dashrightarrow{ \sf{ \sqrt{ {(14)}^{2} + {(1)}^{2} } }}$

$\dashrightarrow{ \sf{ \sqrt{196 + 1}}}$

$\dashrightarrow{ \boxed{ \sf{ \sqrt{197 \:} \: units}}}$

Hope I helped!

Best regards! :D

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3 years ago

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Marrrta [24]

The fireworks display occur at 550.331 ft

The generic equation for a hyperbola provided by: is the best method for solving the problem in this instance.

$\frac{y^{2} }{b^{2} } - \frac{x^{2} }{a^{2} } = 1$ -

We presume that the y-axis is where the hyperbola's longer axis lies.

In this instance, the sound traveled a total of 1100 meters, and the value of a can be calculated as follows:

2a = 1100

a = 550ft

Given that the two people are 3 miles apart, we can calculate the value of c in feet as follows:

1 mile = 5280 feet

∴ 3 miles = 15840 ft = c

And this is how we may determine what b is worth:

$b^{2} = c^{2} - a^{2}$

$b^{2} = 15840^{2} - 550^{2}$

$b^{2} = 250603100$

Our equation is thus given by:

$\frac{y^{2} }{250603100 } - \frac{x^{2} }{550^{2} } = 1$

And for this case we want to find the value of x when y = 3ft=15840ft and solving for x we get,

$\frac{15840^{2} }{250603100 } - \frac{x^{2} }{550^{2} } = 1$

$\frac{15840^{2} }{250603100 } - 1 = \frac{x^{2} }{550^{2} }$

$1.001- 1 = \frac{x^{2} }{550^{2} }$

$0.001 = \frac{x^{2} }{550^{2} }$

$302865.145= {x^{2}$

x = 550.331 ft.

The fireworks display occur at 550.331 ft.

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2 years ago

An agronomist hopes that a new fertilizer she has developed will enable grape growers to increase the yield of each grapevine by

Phoenix [80]

Answer:

Yes. Fertilization increases grape yields by more than 5 pounds.

Step-by-step explanation:

Let f-fertilized and o-old(unfertilized)

#First, we use our data to calculate the standard error:

$SE(\bar y_f-\bar y_o)=\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}\\\\=\sqrt{\frac{3.7^2}{44}+\frac{3.4^2}{47}}\\\\=0.7464$

#State both null and alternative hypothesis:

$H_o:\mu_f-\mu_o=0\\\\H_A=\mu_f-\mu_o>0$

#We determine our degrees of freedom as 87.02(using R), we now compute the t-value as:

$t=\frac{\bar y_f-\bar y_o}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{53.4-52.1}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{1.3}{0.7464}\\\\t=1.7417\\\\P=P(t_{87.0>1.7417}=1-0.9575=0.0.0425$

Since, the p-value is low, we Reject the null hypothesis. The is enough evidence suggesting that grape yields increase by more than 5 pounds than mean yields of unfertilized grapes.

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4 years ago

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docker41 [41]

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3 years ago

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atroni [7]

Answer:

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Step-by-step explanation:

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