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
jenyasd209 [6]
3 years ago
7

Any idea what the answer could be??????

Mathematics
1 answer:
sveta [45]3 years ago
3 0

Here is the answer with an explanation!

You might be interested in
Solve st + 3t = 6 for s
vaieri [72.5K]

st + 3t = 6 for s

Subtract 3t to both sides

st + 3t  - 3t = 6 - 3t

Simplify

st = 6 - 3t

Divide both sides by t

st/t = (6-3t)/3

simplify

s = 6/3 - 3t/3

s = 2 - t


6 0
3 years ago
Which of the following decimal numbers are examples of terminating decimals?
andrew11 [14]

answer:

the answer to the question is c) 14.1

5 0
3 years ago
Read 2 more answers
A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o
o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that \mu = \frac{n}{60}, in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

n = 25, and thus, \mu = \frac{25}{60} = 0.4167

This probability is:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that n = 6

Each of these days, 0.6592 probability of no accident on this stretch, which means that p = 0.6592.

This probability is P(X = 2). So

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

P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

\mu = 0.6\frac{n}{60} = 0.01n

80 miles:

This means that n = 80. So

\mu = 0.01(80) = 0.8

The probability of no damaging accidents is P(X = 0). So

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0
3 years ago
(5у — 4)°<br> Зу<br> (2x + 13)°
ale4655 [162]

Answer:

No Solution.

Step-by-step explanation:

Equality equation:

— 2y – Зу +8 = 8 — 5у – 12  (Given)

-5y + 8 = -4 - 5y (distributive property of equality)

-5y = -12 - 5y (subtraction property of equality)

= -12 (addition property of equality)

when you add 5y on both sides it cancels the 5y out, it's no solution.

7 0
3 years ago
HARDEST MATH QUESTION IN EXISTENCE. CAN YOU ANSWER?
Svetach [21]

Answer:

The answer is 1.2

Step-by-step explanation:

u add 2.2 and 1.2 which you get 3.4. The difference is 1.2.

8 0
3 years ago
Other questions:
  • Please help: solve for x
    11·1 answer
  • Jenny wants a total of 12 barbies dolls by the time she is 9 years old. Right now she has 4 barbie dolls. How many more does she
    6·2 answers
  • Which angle is an adjacent interior angle to ∠JKM? ∠JKL ∠MKL ∠KLM ∠LMK
    8·2 answers
  • Compare the functions below.
    6·2 answers
  • What is a rule for the sequence 1.8, 2.85, 3.90, 4.95, 6
    5·1 answer
  • Will someone plz give me brainliest :(
    9·2 answers
  • Explain how you know that 4/5 =8/10
    10·1 answer
  • Given f() = 3x – 29<br> Find f (4)
    13·2 answers
  • I need help about this ética ofnthe fillowing​
    6·1 answer
  • Pls help!!!! Urgent pls answer 19-36
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!