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Free_Kalibri [48]
3 years ago
11

At a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 12 m. T

he inner edge of the sidewalk is a circle with a radius of 8 m.
Write and simplify an expression for the exact area of the sidewalk.

Find the approximate area of the sidewalk. Use 3.14 to approximate π.

Mathematics
1 answer:
cestrela7 [59]3 years ago
5 0

Answer:

The answer would be 251.2 for the side walk if the radius for the whole circle is 12 and the small circle's radius would be 8. Only for the answer with the 8 m and 12 m radius NOT the 11m and 9 m

The equation would be:

144 x 3.14 - 64 x 3.14 = 251.2

Step-by-step explanation:

First, you need to find out what the whole circle is:

The whole circle is 452.16

Then, find the inner circle :

8 x 8 x 3.14  

200.96

Last, subtract the whole by the inner:

452.16 - 200.96 = 251.2

hope this helps!

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write the equation in slope intercept form of the line that passes through -1, 11 and is parallel to the graph of y = -8x + 3
lawyer [7]
<h2>Answer: The line from the question [ y = -8x + 3 ] passes through the point ( -1, 11 ). </h2>

<h3 /><h3>Step-by-step explanation: </h3>

<u>Find the slope of the parallel line</u>

When two lines are parallel, they have the same slope.

                 ⇒  if the slope of this line = - 8

                      then the slope of the parallel line (m) = - 8

<u>Determine the equation</u>

We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:  

                               ⇒  y - 11  =  - 8 (x - (-1))                                

                                ∴  y - 11  = - 8 (x + 1)

We can also write the equation in the slope-intercept form by making y the subject of the equation and expanding the bracket to simplify:

                      since   y - 11  = - 8 (x + 1)

                                        y  = - 8 x  + 3

The line from the question [ y = -8x + 3 ] passes through the point ( -1, 11 ).

5 0
2 years ago
Solve for g : 3+2g=5g-9
KiRa [710]
3+2g=5g-9
5g-2g=3+9
3g=12
3g/3=12/3
g=4
5 0
3 years ago
In the Salk vaccine field trial, 400,000 children were part of a randomized controlled double-blind experiment. Just about half
EastWind [94]

Answer:

Step-by-step explanation:

From the given information:

The number of children that were randomly allocated to each vaccination group; n₁ = 200,000

No of polio cases X₁ = 57

Now, in the vaccine group:

the proportion of polio cases is:

\hat p_1 = \dfrac{57}{200000}

=  0.000285

The number of children that were randomly allocated to the placebo group, n₂ = 200,000

No of polio cases X₂ = 142

In the placebo group

the proportion of polio cases is:

\hat p_2 = \dfrac{142}{200000}

Null and alternative hypothesis is computed as follows:

H₀: There is no difference in the proportions of polio cases between both groups.  

H₁: There is a difference in the proportions of polio cases between both groups.

Let assume that the level of significance ∝ = 0.05

The test statistic  can be computed as:

Z = \dfrac{\hat p_1-\hat p_2}{\sqrt{\dfrac{\hat p_1 \hat q_1}{n_1}+ \dfrac{\hat p_2 \hat q_2}{n_2}}}

Z = \dfrac{0.000285-0.000710}{\sqrt{\dfrac{0.000285(1-0.000285)}{200000}+ \dfrac{0.000710(1-0.000710)}{200000}}}

Z = \dfrac{-4.25\times 10^{-4}}{\sqrt{\dfrac{0.000285(0.999715)}{200000}+ \dfrac{0.000710(0.99929)}{200000}}}

Z = - 6.03

P-value =  2P(Z < -6.03)

From the Z - tables

P-value =  2 × 0.0000

= 0.000

We reject the H₀ provided that P-value is very less.

Therefore, we may conclude that there is a difference in the proportions of polio cases between the vaccine group and placebo group not due to chance.

7 0
2 years ago
6.) You are standing 50 meters from a hot air balloon that is
lawyer [7]

Answer:

26.6 meters

Step-by-step explanation:

tan = opp/adj

tan28 = opp/50

50 * tan28 = opp

26.59 = opp

4 0
2 years ago
What is the dependent variable
Olenka [21]
Current. This is because the current is dependent on the light intensity which is an independent variable. The current is also affected by the light intensity. 
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