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3 years ago

The legs of a right triangle have lenghts 10 and 24. what is the length of the hypotenuse?

1 answer:

6 0

Use the pythagorean theorem
a^2 + b^2 = c^2
the a and b values are the legs and the c value is the hypotenuse

so in this case you would do:
10^2 + 24^2 = c^2
100 + 576 = c^2
c = square root of 676
c = 26

hope this helps :)

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

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.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

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

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

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

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

3 years ago

Hey can you please answer this question on 7th grade exponents? thank you! boo

Ket [755]

A. 75 x 0.0016= .12 Has a value between 0 and 1

B. 3^5 / 3^-6 = 177147 Does not a value between 0 and 1

C. 1/4^3 x 1/4^2 = about 0.0009 Has a value between 0 and 1

D. -7^5 / -7^7 = about 0.02 Has a value between 0 and 1

8 0

4 years ago

Read 2 more answers

Find the measure of the indicated angle to the nearest degree

hjlf

Answer:

28°

Step-by-step explanation:

sin ? = 19/41

? = arcsin (19/41)

? = 28° (rounded to the nearest degree)

Answered by GAUTHMATH

8 0

3 years ago

You have 40 different books (20 math books, 15 history books, and 5 geography books). a. You pick one book at random. What is th

svetoff [14.1K]

Answer:

a. $\frac{3}{8}$

b. $\frac{7}{52}$

c. $\frac{95}{156}$

Step-by-step explanation:

Given,

Total books = 40,

Maths books = 20,

History books = 15,

Geography books = 5,

a. If a book is randomly selected,

Then the probability that the book is a history book,

$=\frac{\text{Ways of choosing history book}}{\text{Ways of choosing any one book}}$

$=\frac{^{15}C_1}{^{40}C_1}$

$=\frac{15}{40}$

$=\frac{3}{8}$

b. If two book is randomly selected,

Then the probability that both books are from history,

$=\frac{\text{Ways of choosing two history book}}{\text{Ways of choosing any two book}}$

$=\frac{^{15}C_2}{^{40}C_2}$

$=\frac{\frac{15!}{2!13!}}{\frac{40!}{2!38!}}$

$=\frac{105}{780}$

$=\frac{7}{52}$

c. If two book is randomly selected,

Ways of selecting any two books from different subjects

= maths and history + maths and geography + history and geography,

$=^{20}C_1\times ^{15}C_1+^{20}C_1\times ^{5}C_1+^{15}C_1\times ^{5}C_1$

$=20\times 15 + 20\times 5 + 15\times 5$

$=300 + 100 + 75$

$=475$

Then the probability that both books are from different subjects

$=\frac{\text{Ways of selecting any two books from different subjects}}{\text{Ways of choosing any two book}}$

$=\frac{475}{780}$

$=\frac{95}{156}$

7 0

3 years ago

Evaluate B2 for B = -4.

andreyandreev [35.5K]

(-4)^2 = 16
thats the answer

8 0

4 years ago

Read 2 more answers