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Artemon [7]
3 years ago
9

A cyclist pedals up a hill that has a 13 degree angle. if the cyclist pedals up the hill for 10 kilometers what is her change in

elevation, x?
A. 022 kilometers
B. 2.2 kilometers
C. 9.7 kilometers
D. 45.0 kilometers
Mathematics
1 answer:
jeka57 [31]3 years ago
7 0
Sinα=h/d

h=dsinα, since d=10km and α=13

h=10sin13 km

h≈2.24951km

h≈2.2 km (to nearest tenth of a kilometer)
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I need help asap !!!!!!
OLEGan [10]

Answer:

B. ZJ = 85

C. KH = 68

Step-by-step explanation:

Given that ZK, ZL, ZM are perpendicular bisectors of ∆GHJ, they form a circumcenter which is equidistant from the vertices of ∆GHJ. According to the circumcenter theorem, ZH = ZG = ZJ = 85.

Also each median formed is equal. This means: KH = GH, LH = LJ, MG = MJ

Since HG = 136, KH = ½ of 136 = 68.

The two options that are true are:

B. ZJ = 85

C. KH = 68

4 0
3 years ago
ANSWER ASAP
Studentka2010 [4]
A) 3
71 fits into 211 three times
4 0
4 years ago
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It would take an apprentice house painter 1.5h longer than his supervisor to paint an apartment. If they work together they can
never [62]
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4 0
3 years ago
For flights from a particular airport in January, there is a 30 percent chance of a flight being delayed because of icy weather.
NeX [460]

Answer:

0.335

Step-by-step explanation:

1. There is a 30 percent chance of a flight being delayed because of icy weather ,then the probability of being delayed is 0.3 and of being not delayed is 0.7.

2. If a flight is delayed because of icy weather, there is a 10 percent chance the flight will also be delayed because of a mechanical problem, then the probability of being delayed is 0.1 and the probabilty of not being delayed is 0.9.

3. If a flight is not delayed because of icy weather, there is a 5 percent chance that it will be delayed because of a mechanical problem (MP), then the probability of being delayed is 0.05 and the probabilty of not being delayed is 0.95. (See attached probability tree)

Delayed of icy weather - 0.3

Delayed of MP when weather is not icy - 0.7·0.05=0.035

Now, if one flight is selected at random from the airport in January, the probability that the flight selected will have at least one of the two types of delays is

0.3+0.035=0.335

HHHHHHHHHHHHHHHHHHHHHEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPP MMMMMMMMMMMMMMMMMEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE NNNNNNNNNNNNNNNNNNNNNNNNOOOOOOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWWWWWWWWW PPPPPPPPPPPPPPPPPPPPPPPPPLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEAAAAAAAAAAAAAAAASSSSSSSSSSSSSSSSSSSSSEEEEEEEEEEEEEEEEE...............

5 0
3 years ago
A large random sample was taken of body temperatures of women at a university. The data from the sample were normally distribute
ivanzaharov [21]

The best estimate of the proportion of all women at this university who have a body temperature more than 2 standard deviations above the mean is 2.28%.

Since the sample is normally distributed and has a mean μ = 98.52 F and a standard deviation, σ = 0.727 F and we need to find the percentage of all womenat this university who have a body temperature more than 2 standards deviations above the mean.

<h3>The normal distribution</h3>

Since the sample is normally distributed, 50% of the sample is below the mean. We have 34% of the sample at 1 standard deviation away from the mean and 47¹/₂% at 2 standard deviations above the mean.

<h3>Percentage below 2 standard deviations away from mean</h3>

So, the percentage below 2 standard deviations from the mean is 50% + 47¹/₂% = 97¹/₂%.

<h3>Percentage above 2 standard deviations away from mean</h3>

So, the percentage above 2 standard deviations from the mean is 100% - 97¹/₂% = 2¹/₂% = 2.5 %

Since 2.28 % is the closest to 2.5 % from the options, the best estimate is 2.28 %.

So, the best estimate of the proportion of all women at this university who have a body temperature more than 2 standard deviations above the mean is 2.28%.

Learn more about normal distribution here:

brainly.com/question/25800303

6 0
2 years ago
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