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Hatshy [7]
2 years ago
13

Solve the following system of equations.

Mathematics
2 answers:
DIA [1.3K]2 years ago
7 0
2x+y-3=0,x-2y=1 okay that's the answer good luck
Elanso [62]2 years ago
5 0
Subtract y from both sides
now we have
2x = -y + 3
divide both sides by 3
x= -y/2 + 3/2
the other x equals 2y +1, set these equal to eachother

2y + 1 = -y/2 + 3/2

add y/2 to both sides

2.5y + 1 = 3/2

subtract 1 from both sides

2.5y = 0.5

divide both sides by 2.5

y=0.2

plug in 0.2 for y in an original equation

x = 2(0.2) + 1

solve

x = 1.4

ANSWER:
x= 1.4
y= 0.2
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PLEASE HELP WITH FINDING AREA
AnnyKZ [126]

When you have an irregular figure, you can divide it so that it becomes multiple normal figures. You already did this in the picture, now you just have to find the area of each shape and then add them all together.

From bottom to top:

A=1/2 (b1+b2)h

A= 1/2 (4+2.24)(2)

A= 6.24

(you have two of these on either side)

A=bh

A=6(4)

A=24

A=1/2 bh

A= 1/2 (4)(4)

A=8

(NOTE: there is an error in your drawing. The height of the triangle cant be determined because the 10 value spans over the entire figure when the wings of the rocket are 4 units long and then the body is 6 units long, making it equal 10 already. I'm assuming that the 10 is supposed to go from the bottom of the rectangle to the top of the triangle)

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6 0
3 years ago
Suppose that each observation in a random sample of 100 fatal bicycle accidents in 2015 was classified according to the day of t
liraira [26]

Answer:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0:  proportion of fatal bicycle accidents in 2015 was the same for all days of the week

against the claim

Ha:  proportion of fatal bicycle accidents in 2015 was not the same for all days of the week

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (ni - npi)²/ npi

which has an approximate chi square distribution with ( n-1)=7-1=  6 d.f

4) The critical region is χ² ≥ χ² (0.05)6 = 12.59

5) Calculations:

χ²= ∑ (16- 14.28)²/14.28 + (12- 14.28)²/14.28 + (12- 14.28)²/14.28 + (13- 14.28)²/14.28 + (14- 14.28)²/14.28 + (15- 14.28)²/14.28 + (18- 14.28)²/14.28

χ²= 1/14.28 [ 2.938+ 5.1984 +5.1984+1.6384+0.0784 +1.6384+13.84]

χ²= 1/14.28[8.1364]

χ²= 0.569= 0.57

6) Conclusion:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

b.<u> It is r</u>easonable to conclude that the proportion of fatal bicycle accidents in 2015 was the same for all days of the week

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3 years ago
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4 years ago
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vovangra [49]

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6 0
3 years ago
Read 2 more answers
A rectangular prism has a height of 12 cm and a square base with sides measuring 5 cm. A pyramid with the same base and the heig
Ksju [112]
The answer is 200 cm³


The volume of the rectangular prism (V1) is:
V1 = l · w · h                       (l - length,  w - width,  h - height)
It is given:
h = 12 cm
w = l = 5 cm (since it has a square base which all sides are the same size).
Thus: V1 = 12 · 5 · 5 = 300 cm³

The volume of pyramid (V2) is:
V2 = 1/3 · l · w · h                   (l - length,  w - width,  h - height)
It is given:
h = 12 cm
w = l = 5 cm (since it has a square base which all sides are the same size).
V2 = 1/3 · 12 · 5 · 5 = 1/3 · 300 = 100 cm³


The volume of the space outside the pyramid but inside the prism (V) is a difference between the volume of the rectangular prism (V1) and the volume of the pyramid (V2): 
V = V1 - V2 = 300 cm³ - 100 cm³ = 200 cm³
3 0
3 years ago
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