All categories

2 years ago

Please answer and help me on this question!

Mathematics

1 answer:

Nady [450]2 years ago

4 0

<h3>Answer: 10.1 cm approximately</h3>

=====================================================

Explanation:

The double tickmarks show that segments DE and EB are the same length.

The diagram shows that DB = 16 cm long

We'll use these facts to find DE

DE+EB = DB

DE+DE = DB

2*DE = DB

DE = DB/2

DE = 16/2

DE = 8

-------------

Now let's focus on triangle DEC. We just found the horizontal leg is 8 units long. The vertical leg is EC which is unknown for now. We'll call it x. The hypotenuse is CD = 9

Use the pythagorean theorem to find x

a^2+b^2 = c^2

8^2+x^2 = 9^2

64+x^2 = 81

x^2 = 81 - 64

x^2 = 17

x = sqrt(17)

That makes EC to be exactly sqrt(17) units long.

If you follow those same steps for triangle ADE, then you'll find the missing length is AE = 6

---------------

So,

AC = AE+EC

AC = 6 + sqrt(17)

AC = 10.1231056256177

AC = 10.1 cm approximately

You might be interested in

HELP!!!

AfilCa [17]

<span>1.
-x = 2 - 3x + 6
-x + 3x = 8
2x = 8
x = 8/2
x = 4

2.
</span><span>Step 1: 2(3x − 4) = 8 + 2x + 4
Step 2: 5x − 6 = 12 + 2x </span>← wrong
The correct step: <span>Step 2: 6x − 8 = 12 + 2x

3.
</span>The next step is <span>2m = 8</span>

3 0

2 years ago

Read 2 more answers

There are 271 songs on Amy's computer. She

shutvik [7]

Answer:

Step-by-step explanation:

You do 271 ÷9

= 30.11111111

The answer is your whole number (30)

4 0

2 years ago

PLZ ANSWER WILL GIVE BRAINLYEST!!!!

klemol [59]

Answer:

x-intercept = 3.6

y-intercept = -1.2

Step-by-step explanation:

4 0

2 years ago

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

2 years ago

If AC = 52, AB = 8x + 8, and BC = 2x + 4, then the length BC is

Reil [10]

Answer:

Step-by-step explanation:

AC = AB + BC

Since we know what each variable equals, we can write out the equation as shown:

52 = 8X + 8 + 2X + 4

Combining like terms:

52 = 10X + 12

Subtract 12 from both sides:

40 = 10X

Divide by 10:

X = 4

Now that we know X, we can figure out BC

BC = 2(4) + 4

BC = 8 + 4

BC = 12

Therefore BC is 12.

5 0

2 years ago

Please answer and help me on this question!​

Please answer and help me on this question!