(c) 0.34 L = ________________

What is the solution of the system of equations?<br><br> {3x−1/2y=−7/2 x+2y=1

k0ka [10]

3x -1/2y = -7/2
x = -2y + 1

3 (-2y + 1) -1/2y = -7/2
-6y +3 -1/2y = -7/2
-13/2y = -13/2
y = 1

x = -2(1) + 1
x = -1

Answer = (-1, 1)

4 0

3 years ago

1) SHOW WORK: Evaluate f(-2):

babymother [125]

Answer:

- 10 and 60

Step-by-step explanation:

1

To evaluate f(- 2) substitute x = - 2 into f(x)

f(- 2) = (3 × - 2) - 4 = - 6 - 4 = - 10

2

To evaluate f(8) substitute x = 8 into f(x)

f(8) = 8² - 4 = 64 - 4 = 60

4 0

3 years ago

Read 2 more answers

The weight of an object on the moon varies directly as its weight on earth. A person who weights 150.94lb on earth weighs 25.66l

Veronika [31]

109.93 lbs

since the quantities vary directly, then

w(moon) = kw(earth ) ← ( k is the constant of variation )

to find k use the given condition

k = w( moon ) / w( earth ) = $\frac{25.66}{150.94}$

given w(earth) = 218.24

w(moon) = $\frac{25.66(218.24)}{150.94}$ = 109.93

8 0

3 years ago

Write –0.825 as a fraction.

olga2289 [7]

Answer:

OK 825/1000 That is the answer.

8 0

2 years ago

Construct the confidence interval for the population mean mu. c = 0.90, x = 16.9, s = 9.0, and n = 45. A 90% confidence inte

Georgia [21]

Answer:

The 90% confidence interval for population mean is $14.7 < \mu < 19.1$

Step-by-step explanation:

From the question we are told that

The sample mean is $\= x = 16.9$

The confidence level is $C = 0.90$

The sample size is $n = 45$

The standard deviation

Now given that the confidence level is 0.90 the level of significance is mathematically evaluated as

$\alpha = 1-0.90$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the standardized normal distribution table. The values is $Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining critical values for $\frac{\alpha }{2}$ instead of that of $\alpha$ is because $\alpha$ represents the area under the normal curve where the confidence level 1 - $\alpha$ (90%) did not cover which include both the left and right tail while $\frac{\alpha }{2}$ is just considering the area of one tail which is what we required calculate the margin of error