This is geometry work

Find the measure of Angle 8. <br><br> measure of angle 7 = 2x+15<br> measure of angle 8 = 3x

aleksandrvk [35]

I do not understand the context of your question, you need extra information to find this out like a picture or a word problem.
If the angles equal each other you would have to solve for x when angle 7 and 8 equal each other. SO... based on that, x would equal 15. This means measure of angle 7 would equal 45 and measure of angle 8 would equal 45. This may not be true because I have no idea what you are asking

3 0

4 years ago

According to postal regulations, a carton is classified as "oversized" if the sum of its height and girth (the perimeter of its

Amanda [17]

Answer:

V(max) = 8712.07 in³

Dimensions:

x (side of the square base) = 16.33 in

girth = 65.32 in

height = 32.67 in

Step-by-step explanation:

Let

x = side of the square base

h = the height of the postal

Then according to problem statement we have:

girth = 4*x (perimeter of the base)

and

4* x + h = 98 (at the most) so h = 98 - 4x (1)

Then

V = x²*h

V = x²* ( 98 - 4x)

V(x) = 98*x² - 4x³

Taking dervatives (both menbers of the equation we have:

V´(x) = 196 x - 12 x² ⇒ V´(x) = 0

196x - 12x² = 0 first root of the equation x = 0

Then 196 -12x = 0 12x = 196 x = 196/12

x = 16,33 in ⇒ girth = 4 * (16.33) ⇒ girth = 65.32 in

and from equation (1)

y = 98 - 4x ⇒ y = 98 -4 (16,33)

y = 32.67 in

and maximun volume of a carton V is

V(max) = (16,33)²* 32,67

V(max) = 8712.07 in³

3 0

3 years ago

Solve 50q-43=52q-81

ladessa [460]

Move the variables to one side and the constants to the other

50q - 43 = 52q - 81

50q (-52q) - 43 (+43) = 52q (-52q) - 81 (+43)

50q - 52q = -81 + 43

-2q = -38

isolate the q, divide -2 from both sides

-2q/-2 = -38/-2

q = -38/-2

q = 19

hope this helps

8 0

4 years ago

Read 2 more answers

2. Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,00

svlad2 [7]

Using the binomial distribution, it is found that:

a) There is a 0.0501 = 5.01% probability that you need to contact four people.

b) You expect to contact 1.82 students until you find one who lives within five miles of you.

c) The standard deviation is of 1.22 students.

d) There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

e) It is expected that 2.75 students live within five miles of you.

For each student, there are only two possible outcomes. Either they live within 5 miles of you, or they do not. The probability of a student living within 5 miles of you is independent of any other student, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

55% of the students live within five miles of you, thus $p = 0.55$ .

Item a:

This probability is P(X = 0) when n = 3(none of the first three living within five miles of you) multiplied by 0.55(the fourth does live within five miles), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$