Does someone know how to do this i need help answering these questions A-D thanks

Department s had 500 units 60% completed in process at the beginning of the period; 9,000 units completed during the period; and

Delicious77 [7]

With the concept of first in, first out method, then we can use the formula below to solve for the number of equivalent units of production for that period.

number of equivalent units of production

= Total number of units completed during that period (A) – Number of units completed in process at the beginning of the period (B) + Number of units completed at the end of the period (C)

= A – B + C

We know that,

A = 9000 units

So we solve for B and C.

B is 60% of the 500 units, therefore:

B = 0.60 * 500 = 300

C is 30% of the 600 units, therefore:

C = 0.30 * 600 = 180

Substituting the values into the equation:

number of equivalent units of production = 9000 – 300 + 180

number of equivalent units of production = 8880 units

Answer:

A. 8880

3 0

2 years ago

What is one way the Pennsylvania colony was able to maintain peace with the Native American people?

myrzilka [38]

B. Hshejsiaiajsgaiajbehsiabsvahajakajbsjs

5 0

2 years ago

Maths question trigonometry help pleassssee

r-ruslan [8.4K]

Answer:

x=4.5

Step-by-step explanation:

Using sohcahtoa we know tangent is equal to (length of side opposite to the angle ) / (lenth of side adjacent to the angle)

therefore: tan (theta) = (x-3) / (x) = 1 / 3

now cross multiply

3(x-3) = x

now solve for x

3x - 9 = x

-9= -2x

x= 4.5

7 0

2 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

2 years ago

SOMEBODY PLEASE HELP QUICK!!! WILL GIVE BRAINLIEST

Anna71 [15]

Carlos is correct

Since we don't know the length of sides PR and XZ, the triangles can't be congruent by the SSS theorem or the SAS theorem, and since we don't know the measure of angles Y and Q, the triangles can't be congruent by the ASA theorem, the SAS theorem or the AAS theorem. Therefore, Carlos is correct.

Carlos is correct. Since the angles P and X are not included between PQ and RQ and XY and YZ, the SAS postulate cannot be used, since it states that the angle must be included between the sides. Unlike with ASA, where there is the AAS theorem for non-included sides, there is not SSA theorem for non-included angles, so the triangles cannot be proven to be congruent.

3 0

3 years ago