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3 years ago

5

1 lapin a vu 6 éléphants en allant à la rivière. Chaque éléphant a vu 2 singes aller vers la rivière. Chaque singe tient 1 perro

quet entre ses mains.
Combien d'animaux vont-ils vers la rivière ?

1 answer:

forsale [732]3 years ago

4 0

il y a cinq animaux qui entrent dans la rivière

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Check for understanding. <br><br> need help with all 3. PLS HELP

Dominik [7]

1. X=5

2. X=-5

3. Y=10

Step-by-step explanation:

those are the answers

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3 years ago

Miss Garcia an art teacher is buying supplies for her next unit on ceramics. Her 25th six graders are making mugs and jugs estim

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1/2 Answer: 1/2

Step-by-step explanation:

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3 years ago

What has your equivalent to 5/8 pound

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Ten ounces the mass would be 0.625

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3 years ago

Use the data in LAWSCH85 for this exercise. (i) Using the same model as in Problem 4 in Chapter 3, state and test the null hypot

natka813 [3]

Answer:

Step-by-step explanation:

In the model

Log (salary) = B0 + B1LSAT +B2GPA +B3log(libvol) +B4log(cost)+B5 rank+u

The hypothesis that rank has no effect on log (salary) is H0:B5 = 0. The estimated equation (now with standard errors) is

Log (salary) = 8.34 + .0047 LSAT + .248 GPA + .095 log(libvol)

(0.53) (.0040) (.090) (.033)

+ .038 log(cost) – .0033 rank

(.032) (.0003)

n = 136, R2 = .842.

The t statistic on rank is –11(i.e. 0.0033/0.0003), which is very significant. If rank decreases by 10 (which is a move up for a law school), median starting salary is predicted to increase by about 3.3%.

(ii) LSAT is not statistically significant (t statistic ≈1.18) but GPA is very significance (t statistic ≈2.76). The test for joint significance is moot given that GPA is so significant, but for completeness the F statistic is about 9.95 (with 2 and 130 df) and p-value ≈.0001.

6 0

3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when $X = 59.5$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{59.5 - 56}{1.12}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0

3 years ago