Berjabat tangan =66

A six sided number cube is laveled with number from 1 to 6 . If a student tolls the cube 150 times howany times should they expe

andrey2020 [161]

Answer:

75 times

Step-by-step explanation:

6 sided die - 1,2,3,4,5,6

3 of these numbers are even

P(even) = 3/6 = 1/2

1/2 * 150 = 75

7 0

3 years ago

I need help -6x > 30

Debora [2.8K]

Answer:

×< -5

Step-by-step explanation:

-6x>30 reverse the sign

x< -5

7 0

2 years ago

The following graph shows a proportional relationship.

Nina [5.8K]

Answer:

1.5

Step-by-step explanation:

$k=\frac{y}{x}\\\\k=\frac{6}{4}\\\\k=1.5$

8 0

2 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

By the Central Limit Theorem

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

$Z = \frac{20.4 - 18.6}{0.6768}$

$Z = 2.66$

$Z = 2.66$ has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

4 0

2 years ago

If a translation maps zl onto ZK, which of the following statements is true? G HS 1 K НІ is twice the measure of 2JK: GH is twic

murzikaleks [220]

9514 1404 393

Answer:

(d) ∠H ≅ ∠J

Step-by-step explanation:

We already know that ∠G is congruent to itself. If we show (by translation or other means) that ∠I ≅ ∠K, then we know that ΔGHI ~ ΔGJK. The third angle in each triangle will be congruent, too.

∠H ≅ ∠J

_____

The problem is concerned with angles, so the first two answer choices are irrelevant. If two angles are shown congruent, the triangles are congruent by AA similarity, so the third answer choice is incorrect.

6 0

2 years ago

Berjabat tangan =66​

Berjabat tangan =66