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dmitriy555 [2]
3 years ago
5

Help me with 26 ASAP Please show work Random Answers will be moderated! Thanks :)

Mathematics
1 answer:
vfiekz [6]3 years ago
4 0
First, convert the distance of a typical marathon to kilometers.
1 mi = 1.609 km
26.2 miles = 42.165 km

Divide this distance by Allan's average speed to get his average time.

42.165 km / 12 km/hr = 3.514 hours
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A.2

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Solve for v.<br>2 = 11 - 3v<br>v=​
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PLEASEE HELPPP! IT DUE TONIGHT!!!
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Step-by-step explanation:

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Read 2 more answers
HELP!!
tamaranim1 [39]
Question 1: <span>The answer is D. which it ended up being <span>0.9979
</span>
Question 2: </span>The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousand

The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousandth (four decimal places). So being that rounding it off would mean your answer would be = ?

Question 3: <span>Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100. 
a. What percentage of the people taking the test score between 400 and 500?
b. Suppose someone receives a score of 630. What percentage of the people taking the test score better? What percentage score worse?
c. A university will not admit a student who does not score in the upper 25% of those taking the test regardless of other criteria. What score is necessary to be considered for admission? </span>z = 600-450 /100 = .5 NORMSDIST(0.5) = .691462<span><span> z = 400-450 /100 = -.5 NORMSDIST(-0.5) = .30854

P( -.5 < z <.5) = .691462 - .30854 = .3829 Or 38.29%
Receiving score of 630:
z = 630-450 /100 = 1.8 NORMSDIST(1.8) = .9641
96.41% score less and 3.59 % score better
upper 25%
z = NORMSINV(0.75)= .6745
.6745 *100 + 450 = 517 Would need score >517 to be considered for admissions </span><span>
Question 4: </span>The z-score for 45cm is found as follows:</span>
Reference to a normal distribution table, gives the cumulative probability as 0.0099.
<span>Therefore about 1% of newborn girls will be 45cm or shorter.</span>
6 0
4 years ago
Read 2 more answers
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