4) Solve each pair of simultaneous equations.

1 answer:

7 0

4) a. x+y=1–(1)
y=2x-8—(2)
(2) into (1)
x+(2x-8)=1
3x-8=1
3x=1+8
x=9/3
x=3—(3)
(3) into (2)
y=2(3)-8
y=-2

ans x=3, y=-2

b. x+y=19—(1)
y=5x+1—(2)

(2) into (1)

x+(5x+1)=19
6x+1=19
6x=19-1
x=18/6
x=3—(3)

(3) into (2)

y=5(3)+1
y=16

ans x=3, y=16

c.x+y=-2—(1)
y=x-10—(2)

(2) into (1)

x+(x-10)=-2
2x-10=-2
2x=-2+10
x=8/2
x=4–(3)

(3) into (2)
y=4-10
y=-6

ans x=4, y=-6

5) 3x=y—(1)
x=y-16–(2)

(2) into (1)

3(y-16)=y
3y-48=y
2y=48
y=48/2
y=24–(3)

(3) into (2)

x=24-16
x=8

ans x=8, y=24

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Which is greater 10,000ML or 10L of juice

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3 years ago

The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$