1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
satela [25.4K]
2 years ago
10

Rick's Phone Service charges $12.52 for an activation fee plus $6 for every gigabyte (g) used.

Mathematics
2 answers:
oee [108]2 years ago
7 0

Answer:

d

Step-by-step explanation:

same explanation as before just replace the current values with the previous explanation

kakasveta [241]2 years ago
7 0

Answer:

After a quick stint of attempting to read the graphs I would speculate D

You might be interested in
Guys I don't need you to answer the question I just want to know what the math is called
Tju [1.3M]

Answer:

that's the quadratic formula:)

Step-by-step explanation:

4 0
3 years ago
The triangles below are similar.
Simora [160]

Answer:

Forgive me if i'm wrong, but I think it is A D and E

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Find how much money will be in the account after the given number of years. (Assume 360 days in a year) & his interest
horrorfan [7]
I think the answer would be $11,352
5 0
3 years ago
Read 2 more answers
You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte
Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \,  otherwise }} \right.

a)  We calculate  the probability that you need less than 5 minutes to get up:

P(x

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate  the probability that you will need between 8 and 13 minutes:

P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3

Therefore, the probability is P(8<x<13)=0.3.

d)  We calculate the probability that you will be late to each of the 9:30am classes next week:

P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6

You have 9:30am classes three times a week.  So, we get:

P=0.6^3=0.216

Therefore, the probability is P=0.216.

e)  We calculate the probability that you are late to at least one 9am class next week:

P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1

Therefore, the probability is P=1.

3 0
3 years ago
Find the difference: 56.34 - 3.1​
Serjik [45]
It is D you have subtract them
6 0
2 years ago
Other questions:
  • What is the value of 3/7 times .1 divided by 5/21
    11·1 answer
  • Fine the midpoint of the segment with the given points (6,-7) and (-5,1)
    15·1 answer
  • As indicated below, write the equations of the line passing through the point (2, 4) and parallel to the line whose equation is
    15·1 answer
  • One-Step Equations
    5·1 answer
  • 309,099,990 expanded form
    15·2 answers
  • The Jenkin's family decided to go to a concert in the park for their family reunion. A total of 29 family members attended. An a
    14·1 answer
  • Need help I am not good at these​
    9·1 answer
  • Please someone help me with this question. Worth branliest if you get it right. Thank you!
    12·2 answers
  • Evaluate (−3a + 4b)32c − 1 when a = 4, b = 2, and c = 3. <br> Plz Help Its a quiz
    5·2 answers
  • X^4+2x^2-48=0 what is the solution set
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!