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3 years ago

10

Gerry plans to place a 24 -foot ladder against the side of his house. The bottom of the ladder will be 8 feet from the house. Ho

w far up the side of the house will the ladder reach? Answer exactly or round to 2 decimal places.

2 answers:

Nadya [2.5K]3 years ago

5 0

Answer:

x = 22,63 ft

Step-by-step explanation:

The side of the house, the ladder and the ground form a right triangle. The hypotenuse is the ladder ( 24 ft) and the legs are the side of the house (x uknown ) and the other leg, is distance between bottom of the ladder and the side of the house, therefore

Pytagoras Theorem :

H² = L₁² + L₂²

(24)² = (8)² + x²

x² = (24)² - (8)²

x² = 576 - 64 ⇒ x² = 512

x = 22,63 ft

blondinia [14]3 years ago

3 0

Answer:

22.63 ft

Step-by-step explanation:

In the right triangle ABC attached,

AB is the length of the ladder which is the hypotenuse

AC is the distance of the ladder's bottom from the house.

We are to determine how far up the side of the house the ladder will reach.

We apply <u>Pythagoras Theorem</u> to solve this.

$\overline{AB}^2=\overline{AC}^2+\overline{BC}^2\\24^2=8^2+\overline{BC}^2\\\overline{BC}^2=576-64\\\overline{BC}^2=512\\BC=\sqrt{512}=22.63 ft$

The ladder will reach 22.63 ft (correct to 2 decimal places) up the wall of the house.

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Citrus2011 [14]

the answer is 4 and 7

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3 years ago

Julie buys a pair of jeans that cost $39. If they have a 25% discount, what is the sale price of the jeans?

NISA [10]

Answer:

$29.25

Step-by-step explanation:

39 x .25 = 9.75

39 - 9.75 = 29.25

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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

3 years ago

Delegates from 10 countries, including Russia,

ankoles [38]

Answer: 564480 arrangements

Step-by-step explanation:

First, let's consider the French and English delegates as one party because they will sit next to each other.

So total delegates if French and English are always together comes out to be 9 instead of 10.

Now, they are to sit in a row and can have different seats so the total number of ways they can sit is= 2 x 9!= 725760 ways.

Now, when Russian and U.S delegates are not to sit next to each other so consider similarly like before consider them as one unit like French and English. Now they can also change their seats so number of way for them= 2 x 2 x 8!= 161280 ways

To find out when French and English delegates are together and Russain and U.S delegates are not together we subtract the way from the original value.

725760-161280= 564480 way.

4 0

3 years ago

Read 2 more answers

How do I work this out?

zvonat [6]

We have to determine the value of $\sum _{m=9} ^{21} (5m+6)$

= (5(9)+6) + (5(10)+6) +(5(11)+6) + .......... + (5(21)+6)

= 51+56+61+66+ ........ + 111

Since, the common difference is 5, hence this series is in arithmetic progression.

Sum of AP is given by the formula:

$\frac{n}{2}[2a+(n-1)d]$

Since, there are 13 terms.

= $\frac{13}{2}[2(51)+(13-1)5]$

= $\frac{13}{2}[102+60]$

= $\frac{13}{2}[162]$

= $13 \times 81$

= 1053

Therefore, the sum of the series is 1053.

So, Option G is the correct answer.

8 0

3 years ago