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3 years ago

9

Use linear approximation to estimate 1/4.002

1 answer:

Nady [450]3 years ago

3 0

The approximation to estimate will be 0.2498

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Find the area of segment CED given the following information:

Hitman42 [59]

Area of sector CAD = 72 / 360 x pi x 6^2 = 7.2pi = 22.6195 in^2

Therefore, area of segment CED = 22.6195 - 17.18 = 5.44 in^2

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4 years ago

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If Rhonda can type a 1000 word report in 4 hours, and while her friend Betty types on the same report, they finish it in 3 hours

Sever21 [200]

R * 4 = 1000; R = 1000 / 4 = 250
(R + B) * 3 = 1000
3R + 3B = 1000,
3B = 1000 - 3R = 1000 - 750 = 250
B = 250/3

t(B) = 1000 / (250/3) = 12 hours

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3 years ago

Please help me with this question!

Vitek1552 [10]

Answer:

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3 years ago

Statistics from Cornell's Northeast Regional Climate Center indicate that Ithaca, NY, gets an average of 35.4 inches of rain eac

alisha [4.7K]

Answer:

c. approximately 90.07%

Step-by-step explanation:

We need to calculate z-statistic of 30 inches of rain in the normal distribution with an average of 35.4 inches of rain each year, and a standard deviation of 4.2 inches.

z score can be calculated using the formula

z= $\frac{X-M}{s}$ where

X =30 inches of rain

M is the average rain in inches (35.4)

s is the standard deviation (4.2)

using the numbers we get z= $\frac{30-35.4}{4.2}$ ≈ −1,286

Then percentage of years does Ithaca get more than 30 inches of rain is

P(z>-1.286) =1- P(z<-1.286)=1-0.0993 ≈0.9007 or 90.7%

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4 years ago

NO LINKS!! Please help me with this problem

Vera_Pavlovna [14]

Information : The given hyperbola is a horizontal hylerbola with its centre (3 , -5) and one of its focus at (9 , -5) and vertex at (7 , -5) and as we can see that the focus and vertex have same y - coordinates, it must have its Transverse axis on line y = - 5.

Now,

it's vertex is given, I.e (7 , -5)

so, length of semi transverse axis will be equal to distance of vertex from centre, i.e

a = 7 - 3 = 4 units

Now, it's focus can be represented as ;

$\qquad \sf \dashrightarrow \: (3 + ae, - 5 )$

so,

ae + 3 = 9

and we know, a = 4

$\qquad \sf \dashrightarrow \: 4e + 3 = 9$

$\qquad \sf \dashrightarrow \: 4e = 6$

$\qquad \sf \dashrightarrow \: e = \cfrac{3}{2}$

Now, let's find the measure of semi - conjugate axis (b)

$\qquad \sf \dashrightarrow \: {b}^{2} = {a}^{2} ( {e}^{2} - 1)$

$\qquad \sf \dashrightarrow \: {b}^{2} = 16( \frac{9}{4} - 1)$

$\qquad \sf \dashrightarrow \: {b}^{2} = 16( \frac{9 - 4}{4} )$

$\qquad \sf \dashrightarrow \: {b}^{2} = 16( \frac{5}{4} )$

$\qquad \sf \dashrightarrow \: {b}^{2} = 20$

$\qquad \sf \dashrightarrow \: b = \sqrt{20}$

So, it's time to write the equation of hyperbola, as we already have the values of a and b ~

$\qquad \sf \dashrightarrow \: \cfrac{ {(x - h)}^{2} }{ {a}^{2} } - \cfrac{( {y - k)}^{2} }{ {b}^{2} } = 1$

[ plug in the values, and h = x - coordinate of centre, and k = y - coordinate of centre ]

$\qquad \sf \dashrightarrow \: \cfrac{ ({x-3)}^{2} }{ {16}^{} } - \dfrac{ {(y+5)}^{2} }{ { {20} }^{} } = 1$

6 0

2 years ago

Read 2 more answers