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olga55 [171]
3 years ago
8

Sos, paper due tmr ! EASY ALGEBRA QUESTION FOR MANY BUT NOT FOR ME ! HELP ME FILL IN THE BLANKS !

Mathematics
1 answer:
kirill115 [55]3 years ago
8 0
So... for the beginning m = 1 cause your slope is just rising and running one everyone time. This makes the equation
y = x + 2
Then for the next question your y value is determined to when x equals 0. This means that y equals -3. To find the m (slope) divide
\frac{y}{x}
for example
\frac{5}{2}  = 2.5
That would be your slope line.
The two points can be any (x,y) points from the table.
Have fun.
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The dollar value v (t) of a certain car model that is t years old is given by the following exponential function.
galina1969 [7]

Answer:

$ 3820

Step-by-step explanation:

When the car is new, it is 0 year old. It means, t = 0

Initial value = 26,000 (0.84)^0 = 26,000(1)

Initial value = $ 26000

Similarly, when car is 11 year old, t = 11

=> 26000(0.84)¹¹ = 26000(0.14691)

≈ $ 3819.842

≈ $ 3820 (round to near ten)

6 0
3 years ago
Kesha sharpened 4 colors of pencils for her teacher. She sharpened equal amounts of red and green pencils. Then she sharpened tw
Nimfa-mama [501]
Variables are taken from the first letter of the color, ex: green = g.

Given:
r = g
2r = b = y
Total =42

So then:
r+ r + 2r + 2r = 42
6r = 42
r = 7

y = 2r
y = 2(7)
14 of the 42 pencils were yellow.
7 0
3 years ago
ASAP can someone help me solve this problem​
sveta [45]

Answer:

Step-by-step explanation:

x° + y° = 180°

2y° - 0.5x° = 180°

y° = 108°

x° = 72°

3 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
EASY 14 POINTS, PLEASE HELP ASAP.
jenyasd209 [6]
Well really it would have to depend on the flashlight. Is it an led bulb or a regular bulb?
5 0
3 years ago
Read 2 more answers
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