Part A

Solve the system of equations using substitution. 2r – t=7 r – t=1 what does r and t equal, hurry please

borishaifa [10]

Answer:

(r,t)=(0,-1)

:)

7 0

3 years ago

7. (ASV, Exercise 2.20 ) - 3 points. A fair die is rolled repeatedly. Use precise notation of probabilities of events and random

Korvikt [17]

Answer:

0.9646 ; 0.5685

Step-by-step explanation:

Given that :

P(obtaining 3 on a die roll) = 1/6 = 0.1667

Obtaining 3 at most 2 times in five trials

Using binomial distribution formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

n = 5 ;, p = 0.1667

P(x ≤ 2) = P(x = 0) + P(x =1) + P(x = 2)

Using the binomial probability :

P(x ≤ 2) = 0.4019 + 0.4019 + 0.1608

P(x ≤ 2) = 0.9646

B.) using geometric distribution :

(1 - p)^x-1 * p

P ≥ 5) = 1 - P(x ≤ 4) = 1 - p(x = 0) + p(x = 1) + p(x =2) + p(x = 3)

1 - (0.1667 + 0.1157+0.0965+0.0804)

= 0.5685

8 0

2 years ago

Which of the following is a solution for the inequality

DedPeter [7]

X=5.5 i think is the answer

4 0

3 years ago

Read 2 more answers

Fill in the blanks to complete the sentence.

inna [77]

Answer:32%

Step-by-step explanation:

330-250=80

80/250=0.32 or 32%

7 0

2 years ago

According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use

taurus [48]

Answer:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=\frac{120}{300}=0.4$ estimated proportion of college students that have a smart phone

$p_o=0.35$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is >0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.1$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

4 0

2 years ago