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3 years ago

The difference of this equation.

Mathematics

1 answer:

yaroslaw [1]3 years ago

4 0

Answer:

$\frac{23}{(y-8)(y+8)}=\frac{23}{y^{2}-64}$

Step-by-step explanation:

$\frac{3y - 1}{(y-8)(y+8)} -\frac{3}{y+8} = \frac{3y - 1}{(y-8)(y+8)} -\frac{3(y-8)}{(y-8)(y+8)} =\\\\=\frac{3y - 1-3y+24}{(y-8)(y+8)} = \frac{23}{(y-8)(y+8)}=\frac{23}{y^{2}-64}$

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Solve 1 over 36 = 6x−3. x = −5 x = negative 3 over 2 x = 7 over 2 x = 1

vlabodo [156]

6*(-5)-3 = -33
6*(-3/2)-3 = -12
6*(7/2)-3 = 18
6*(1)-3=3

3 0

3 years ago

Read 2 more answers

Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. A consume

Ivan

Answer:

Type I error: Concluding that mean mileage is less than 32 miles per hour when actually it is greater than or equal to 32 miles per gallon.

Step-by-step explanation:

We are given the following in the question:

Hypothesis:

Mean mileage for the Carter Motor Company's new sedan

We can design the null hypothesis and alternate hypothesis as:

$H_{0}: \mu \geq 32\text{ miles per gallon}\\H_A: \mu < 32\text{ miles per gallon}$

Type I error:

It is the false positive error.
It is the error of rejection a true hypothesis.

Type II error:

It is the false negative error.
It is the non rejection of a false null hypothesis.

Thus, type I error for the given hypothesis is concluding that mean mileage is less than 32 miles per hour when actually it is greater than or equal to 32 miles per gallon.

Type II error would be concluding that mean mileage is greater than or equal to 32 miles per gallon when actually it is less than 32 miles per gallon.

3 0

3 years ago

A manager hires labour and rents capital equipment in a very competitive market. Currently the wage rate is GH¢2 per hour and ca

Veseljchak [2.6K]

Solution :

It is given that the manager hires a labor and he rents the capital equipment $\text{in a very competitive market}$ .

Presently the rate of the wage is at $ 10 per hour and the capital is been rented at $ 0.25. If the $\text{marginal product}$ of the labor is 50 units of the output per hour and the marginal.

Therefore, the answer is

14 + 10 = 24

8 0

3 years ago

A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg

polet [3.4K]

Answer:

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

= 0.0355

b. Compute the probability that exactly 4 will withdraw (to 4 decimals).

= 0.1304

c. Compute the probability that more than 3 will withdraw (to 4 decimals).

= 0.8929

d. Compute the expected number of withdrawals.

= 6

Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

$P(X = x) = nCx p^{x} q^{n - x}$

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 1) = 20C1(0.3)^1(0.7)^{19}\\P(X = 1) =20 * 0.3 * 0.001139889518\\P(X = 1) = 0.006839337111$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 0) = 20C0(0.3)^0(0.7)^{20}\\P(X = 0) =1 * 1 * 0.000797922662\\P(X = 0) = 0.000797922662$

Finally, the probability that 2 or fewer students will withdraw is

$P(X = 2) + P(X = 1) + P(X = 0) \\= 0.027845872524 + 0.006839337111 + 0.000797922662\\= 0.035483132297\\= 0.0355$

b) Compute the probability that exactly 4 will withdraw.

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304$

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716$

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1 - 0.1071=0.8929

d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

5 0

3 years ago

Can you guy help for this questing 1234

postnew [5]

Answer:

1. Median

2. Altitude

3. Altitude

4. Neither

Step-by-step explanation:

Median- splits the side into two congruent parts

Altitude- Makes a right angle

1. Splits the side. Median

2. Makes a right angle. Altitude

3. Makes a right angle. Altitude

4. Doesn't make a right angle or split the sides. Neither

4 0

3 years ago