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BigorU [14]
2 years ago
5

Two sides of an obtuse triangle measure 9 inches and 14 inches. The length of longest side is unknown. What is the smallest poss

ible whole-number length of the unknown side? 16 inches 17 inches 24 inches 25 inches
Mathematics
2 answers:
julia-pushkina [17]2 years ago
4 0

Answer:

17 inches

Step-by-step explanation:

An obtuse triangle is the triangle in which one of the side is the longest. It contains an obtuse angle and the longest side is the side that is opposite to the vertex of the obtuse angle.

Let the three sides of the obtuse triangle be a, b and c respectively with c as the longest side. Let a = 9 inches and b = 14 inches.

Now we know that for an obtuse triangle,

$c^2 > a^2 +b^2$

$c^2 > (9)^2 +(14)^2$

$c^2 > 81 +196$

$c^2 > 277$

c > 16.64

Therefore the smallest possible whole number is 17 inches.

True [87]2 years ago
3 0

Answer:

B)17

Step-by-step explanation:

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Answer:

The answer is C. Lila made an error in Step 3 when she did not use the x- and y-coordinates from the same ordered pair.

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5 0
2 years ago
Read 2 more answers
The engineers designing the All Aboard Railroad between Boca Raton and Jupiter decide to create parallel tracks through this por
olga2289 [7]

Answer:

there are no signs between the x and y and constant

it could be

2x+5y=15

2x+5y=-15

-2x+5y=15

2x-5y=15

for ax+by=c, the equation of a line paralell to that is

ax+by=d where a=a, b=b, and c and d are constants

(for this answer, I'm going to use 2x+5y=15)

given 2x+5y=15, the equation of a line paralell to that is 2x+5y=d

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2x+5y=d

2(4)+5(-2)=d

8-10=d

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5 0
2 years ago
A helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. Angle ABC
In-s [12.5K]

Please consider the attached graph.

We have been given that a helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. The measure of angle ABC is 150°. We are asked to find the area of triangle ABC.

We will use trigonometric area formula to solve our given problem.

A=\frac{1}{2}a\cdot c\cdot\text{sin}(b), where angle b is angle between sides a and c.

For our given triangle a=5,c=8 and measure of angle b is 150 degrees.

A=\frac{1}{2}(5)\cdot (8)\cdot\text{sin}(150^{\circ})

A=5\cdot 4\cdot(0.5)

A=10

Therefore, the area of the triangle ABC is 10 square kilo-meters and option 'c' is the correct choice.

6 0
2 years ago
The Acme Candy Company claims that​ 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers
vichka [17]

Answer:

Yes, it would be statistically significant

Step-by-step explanation:

The information given are;

The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%

Number of jawbreakers in the sample, n = 800

The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = \mu _ {\hat p} =p

The formula for the standard deviation of a proportion is \sigma  _{\hat p} =\sqrt{\dfrac{p(1-p)}{n} }

Solving for the standard deviation gives;

\sigma  _{\hat p} =\sqrt{\dfrac{0.6 \cdot (1-0.6)}{800} } = 0.0173

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4  in the sample of 800 = 800*0.6 = 480

For statistical significance the difference from the mean = 2×\sigma _{\hat p} = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7

The z-score of 494 jawbreakers is given as follows;

Z=\dfrac{x-\mu _{\hat p} }{\sigma _{\hat p}  }

Z=\dfrac{494-480 }{0.0173  } = 230.94

Therefore, the z-score more than 2 ×\sigma _{\hat p} which is significant.

8 0
2 years ago
The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 0
2 years ago
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