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2 years ago

A right triangle has a leg of 17 cm and a hypotenuse of 25 cm. What is the length of the other leg? Round to the nearest tenth.

Mathematics

2 answers:

balu736 [363]2 years ago

7 0

Answer:

24 feet.

Step-by-step explanation:

The Pythagorean theorem states that the sum of the squares of the two legs is equal to the hypotenuse squared.

7² + x² = 25² → I'll use "x" to represent the unknown leg.

49 + x² = 625

x² = 576

x= ± 24 → Eliminate the -24 answer choice; a side length cannot be negative.

The other leg is 24 ft.

~hope this helps~

Rudik [331]2 years ago

3 0

Answer:

18.3

Step-by-step explanation:

Pythagorean Theorem = a^2 + b^2 = c^2

a, b = legs

c = hypotenuse

17^2 + b^2 = 25^2

b = 18.3

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Natalija [7]

Answer:

Domain and Range of g(f(x)) are 'All real numbers' and {y | y>6 } respectively

Step-by-step explanation:

We have the functions, f(x) = eˣ and g(x) = x+6

So, their composition will be g(f(x)).

Then, g(f(x)) = g(eˣ) = eˣ+6

Thus, g(f(x)) = eˣ+6.

Since the domain and range of f(x) = eˣ are all real numbers and positive real numbers respectively.

Moreover, the function g(f(x)) = eˣ+6 is the function f(x) translated up by 6 units.

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3 years ago

Read 2 more answers

Consider the two number sets. Which set has the higher mean and what is that mean?

Otrada [13]

The mean is where you add all of the numbers in that set together, and divide it by the amount of numbers you have. E.g. For Set 1 - 10 + 15 + 20 + 25 + 30 + 50 = 150 ÷ 6 = 25

For Set 2 you just repeat the process:

1. Add the numbers in the set together, this gives you a total of 111

2. Then divide 111 by 5 as you have 5 numbers, giving you 22.2

Therefore set 1 has the higher mean :)

The answer is C, set 1, 25 :)

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3 years ago

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

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3 years ago

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3 years ago

Mr. Smith is ordering steaks to serve at a poker game with his friends. The first company charges a delivery fee of $10 plus $15

gregori [183]

Answer:

he would have to order 3 steaks from the first company and 3 from the second

Step-by-step explanation:

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