Can somebody please help

Find 3 ordered pairs which are Solutions to the equation 2x - 3y = 12

KonstantinChe [14]

2x-3(0)=12
2x=12
x=6
(6,0)

2(0)-3y=12
-3y=12
y=-4
(0,-4)

2x-3(2)=12
2x-6=12
2x=18
x=9
(9,2)

ANSWERS: (6,0) (0,-4) (9,2)

7 0

3 years ago

Help please! show work thanks

Andre45 [30]

Answers
b = 2.77 m
A = 43.0°
C = 111.1°

cosine law to find b

$b^2 = a^2 + c^2 -2ac \cos B \\ \\
b = \sqrt{a^2 + c^2 -2ac \cos B} \\ \\
b = \sqrt{4.33^2 + 5.92^2 - 2(4.33)(5.92) \cos 25.9} \\
b = 2.7708\ m$

b = 2.7708\ m

Find angle A with sine law

$\displaystyle
\frac{\sin A}{a} = \frac{\sin B}{b} \\ \\
\sin A = \frac{a \sin B}{b} \\ \\
A = \sin^{-1} \left[ \frac{a \sin B}{b} \right] \\ \\
A = \sin^{-1} \left[ \frac{4.33 \sin 25.9}{2.7708} \right] \\ \\
A = 43.0467020$

Find C with angles in triangle sum to 180

A + B + C = 180
C = 180 - A - B
C = 180 - 43.0467020 - 25.9
C = 111.1

6 0

3 years ago

A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th

Burka [1]

Answer:

Confidence interval : $21.506$ to $24.493$

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = $\bar{x}=\frac{\sum x}{n}$

Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$