<u>First problem :</u>
probability to draw a red paper : 42/(42+18)=7/10

In 110 tries one can thus expect to draw 7/10*110=77 red papers

<u>Second problem :</u>
Since for 150 trials (which will give the least error) we have got 135 yellow balls, we can expect twice as much in 300=150*2 trials, which makes 270 yellow balls.

3 0

3 years ago

Trisha is taking care of the Han family's dogs.The Hans leave 7 cans of dog food for the 3days they'll be away.How much food wil

Verdich [7]

Answer:

2 1/3 food

Step-by-step explanation:

Cans of dog food = 7

Number of days = 3

If Trisha feeds them an equal amount each day, then each day the dogs will get:

$\frac{7}{3}=2\frac{1}{3}$ food

7 0

2 years ago

The CEO of a large electric utility claims that more than 75 percent of his customers are very satisfied with the service they r

vichka [17]

Answer:

$z=\frac{0.77-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}=0.462$

Now we can find the p value using the alternative hypothesis with this probability:

$p_v =P(z>0.462)=0.322$

Since the p value is large enough, we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%

Step-by-step explanation:

Information provided

n=100 represent the random sample selected

$\hat p=0.77$ estimated proportion of students that are satisfied

is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if more than 75 percent of his customers are very satisfied with the service they receive, then the system of hypothesis is.:

Null hypothesis: $p\leq 0.75$

Alternative hypothesis: $p > 0.75$

The statistic is given by:

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

Replacing the info given we got:

$z=\frac{0.77-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}=0.462$

Now we can find the p value using the alternative hypothesis with this probability:

$p_v =P(z>0.462)=0.322$

Since the p value is large enough we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%

4 0

3 years ago