Solve. -16 - 8 A. 24 B. 8 C. -8 D. -24

Sam bought a jacket for $17, which is one-sixth of the original price. How much did the jacket cost originally?

kkurt [141]

Answer:

=102

Step-by-step explanation:

17=1/6 of the price

x=1/1 of the price

x=1*17*6=102

7 0

2 years ago

What is 2 + 5 - 2 x 9 = ? The person who gets it right, I will make you brainliest.

Minchanka [31]

PEMDAS
Multiplication: 2 + 5 - 18=
Addition: 7-18=
Subtraction: -11
Answer=-11

5 0

3 years ago

Read 2 more answers

I really need help with this answer can someone help?

valina [46]

Answer:

$7075 or 7718

Step-by-step explanation:

91100-6200=84900

84900/11= 7718.18181818 or rounded= 7718

(This is if the december month doesnt count.)

84900/12=7075

(If december is included.)

7 0

3 years ago

Read 2 more answers

Conduct the appropriate hypothesis test and compute the test statistic. A company that produces fishing line undergoes random te

andre [41]

Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 pound

Sample mean, $\bar{x}$ = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 2 pounds

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}$

We use one-tailed(left) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29.1 - 30}{\frac{2}{\sqrt{20}} } = -2.012$

Now,

$t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01

5 0

3 years ago

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

3 years ago