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Soloha48 [4]
3 years ago
13

A 15 foot pole extends t feet below ground and 10 feet above ground. Choose the linear equation that models the situation.

Mathematics
1 answer:
My name is Ann [436]3 years ago
8 0

No options are given for the question :

Answer:

t + 10 = 15

Step-by-step explanation:

Given that :

Length of pole = 15 feet

Height below the ground = t

Height above the ground = 10

Tge linear equation which represents the scenario can be modeled as :

Entire length of pole = 15 feets

Length above ground = 10

Length below ground = t

Length below + length above = total length

t + 10 = 15

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If 1/8(8x+15)24, what is the value of x?
Lelechka [254]

Answer:

x=22.125  or  x=\frac{177}{8}

Step-by-step explanation:

<u><em>The correct equation is</em></u>

1/8(8x+15)=24

solve for x

Multiply by 8 both sides to remove the fraction

(8x+15)=192

subtract 15 both sides

8x=192-15\\8x=177

divide by 8 both sides

x=\frac{177}{8}

x=22.125

3 0
3 years ago
Look at this plz
Ivan

Answer:

gosh...…. dang brainly why do you have to be like that. you were literally trying to help ppl hippo and they just shut you down like that... ugh why rude.

Step-by-step explanation:

im sorry :(

5 0
3 years ago
Suppose you create a graph of the cost function, C = 20n + 500 of a new
ASHA 777 [7]
The answer is C. loss section
3 0
3 years ago
What is the hypothesis in this conditional statement?
nexus9112 [7]
The answer is
A. 
Hope This Helps

8 0
3 years ago
A random sample of 77 fields of corn has a mean yield of 26.226.2 bushels per acre and standard deviation of 2.322.32 bushels pe
PSYCHO15rus [73]

Answer:

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

n = 77

mean u = 26,226.2  bushels per acre

standard deviation s = 2,322.32

let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let  A =  (u - E) *  (\sqrt{n}  / s)   be the test statistic

we want      P( average_l <  A  < average_u )  = 95%

look for  lower 2.5%  and the upper 97.5%  Because I think this is a 2-tail test

average_l =  -1.96  which corresponds to the 2.5%

average_u = 1.96

P(  -1.96  <  A  <  1.96)  =  95%

P(  -1.96  <  (u - E) *  (\sqrt{n}  / s)  <  1.96)  =  95%

Solve for the true mean E ok

E   <   u + 1.96* (s  / \sqrt{n})

from  -1.96  <  (u - E) *  (\sqrt{n}  / s)

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s  / \sqrt{n})  < E

26,226.2 -  518.7197348105429466  < E

25,707.48026519  < E

lower bound is 25,707.48026519

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

7 0
3 years ago
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