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3 years ago

Does anyone know a rule that work for all of these

Mathematics

1 answer:

Oxana [17]3 years ago

3 0

Divide each term in the y column by 2 to go from this list {8, 2, 0, 2, 8} to this list {4, 1, 0, 1, 4}

This list {4, 1, 0, 1, 4} is a bunch of perfect squares so it suggests that y = x^2

However we must double the values to get back to the original list. So the rule is y = 2*x^2

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2 years ago

At a 20th high school reunion, all the classmates were asked the number of children they had. The probability of having a partic

finlep [7]

Answer:

a) We need to check two conditions:

1) $\sum_{i=1}^n P_i = 1$

$0.05+0.14+0.34+0.24+0.11+0.07+0.02+0.02+0.01= 1$

2) $P_i \geq 0 , \forall i=1,2,...,n$

So we satisfy the two conditions so then we have a probability distribution

b) $P(C \geq 1)$

And we can use the complement rule and we got:

$P(C \geq 1)= 1-P(C$

c) $P(C=0) = 0.05$

d) For this case we see that the result from part b use the probability calculated from part c using the complement rule.

Step-by-step explanation:

For this case we have the following probability distribution given:

C 0 1 2 3 4 5 6 7 8

P 0.05 0.14 0.34 0.24 0.11 0.07 0.02 0.02 0.01

And we assume the following questions:

a) Verify that this is a probability distribution

We need to check two conditions:

1) $\sum_{i=1}^n P_i = 1$

$0.05+0.14+0.34+0.24+0.11+0.07+0.02+0.02+0.01= 1$

2) $P_i \geq 0 , \forall i=1,2,...,n$

So we satisfy the two conditions so then we have a probability distribution

b) What is the probability one randonmly chosen classmate has at least one child

For this case we want this probability:

$P(C \geq 1)$

And we can use the complement rule and we got:

$P(C \geq 1)= 1-P(C$

c) What is the probability one randonmly chosen classmate has no children

For this case we want this probability:

$P(C=0) = 0.05$

d) Look at the answers for parts b and c and explain their relationship

For this case we see that the result from part b use the probability calculated from part c using the complement rule.

5 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>