Answer:
Por ejemplo, podrías utilizar la división para determinar cómo repartir 40 ... 15 ÷ 5 = 3. Podrías también usar la recta numérica para modelar ésta división. ... ¿Cuáles de las siguientes expresiones representan dividir $56 en partes iguales entre 7 ... “5, 10, 15, 20, 25, 30. Debo saltar 6 veces para llegar a 30.” 30 ÷ 5 = 6.
Step-by-step explanation:
Answer:
1.50 for a water
2.80 for one nacho
Step-by-step explanation:
3w + 2n = 9.5
2w + 1n = 5.5
<em>good luck, i hope this helps :)
</em>
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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 thousandThe 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>
