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alukav5142 [94]

3 years ago

7

The lengths of two sides of a right triangle are 5 inches and 8 inches. What is the difference between the two possible lengths

of the third side of the triangle? Round your answer to the nearest tenth.

1 answer:

scZoUnD [109]3 years ago

3 0

If 5 in and 8 in are legs:
hypotenuse = √(5²+8²) = √89 ≈ 9.4 in.

If 5 in is the leg and 8 in is the hypotenuse:
other leg = √(8²-5²) = √39 ≈ 6.2 in.

<span>The difference between the two possible lengths of the third side =
9.4 - 6.2 = 3.2 in.</span>

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What is the approximate perimeter of a semicircle with a radius of 8 m? Use pi star times = 3.14. Express your final answer to t

irga5000 [103]

The perimeter of a semicircle consists of two parts. (the curve and bottom)
That curve is half the distance around the circle, since it's been split in half.
The distance around a circle, the circumfrence, is equal to 2πr, where r is the radius of that circle. In this case, the circumfrence of the entire circle would be 16π. and so that curve would have a length of just 8π.
Using 3.14 for π, 8π = 8×3.14 = 25.12.
As for the flat part, that is the diameter (distance across) our circle.
The radius is the distance from the center of a circle to its edge, and always has half the length of the diameter. (you can break the diameter down into two radii)
If our radius is 8 meters, our diameter (the flat part of that semicircle) must be 16.
Now we add up the two parts of the perimeter...25.12 + 16 = 41.12.

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3 years ago

Identify the variable that is continuous data used discretely. Select all that apply. There are 3 bugs. The bugs are red, yellow

Nata [24]

Answer:

The age of each bug

Step-by-step explanation:

Discrete variables are variables which are counted and can only take on whole number values such as 1, 2, 3...,. For example, the number of students in a class; the number of legs of a spider.

Continuous variables are variables that are measured and can take on a whole range of values such as 1.0, 1.1, 1.2...,. For example, the wingspan of a bug, the length of leaves; the exact age of students in a class.

Though age is a continuous variable if the exact age is required as it can take up a wide range of values such as 2.5 years; 2.3 years, 5.8 days etc,. However, in some instances, it can be rounded up to whole number values in years, months, weeks or even days.

For example, the age of a bug in days could be 1 day, 2 days, or 14 days.

The age of a student who is 12.2 years can be given as 12 years.

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3 years ago

Marsha recorded the time it took seven children of different ages to run one lap around the track.

max2010maxim [7]

The answer is C. 205.

7 0

2 years ago

Read 2 more answers

A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January

Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

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.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

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

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

$E(X) = np = 10*0.23 = 2.3$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

The expected number is 2.3 and the standard deviation is 1.33.

5 0

3 years ago

The Sanchez family is planning a party. They want to budget spending $25 per guest. They already spent $200 on supplies and will

OverLord2011 [107]

Answer:

4

Step-by-step explanation:

20×25 = 500

20 guest

25 dollars per guest

75s+200= 500

75s = 300

S = 4

They can afford to buy 4 cases of snacks and stay within their budget.

3 0

2 years ago