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AleksAgata [21]
3 years ago
5

The mean weight of the packages Joan shipped was 2.5 pounds. If Joan mailed four packages and three of them had weights of 1.7,

3.2, and 2.8 pounds, then what did the other package weigh?
Mathematics
2 answers:
Blababa [14]3 years ago
8 0
The mean weight of the packages is 2.5, so the total weight of the packages is found by multiplying 2.5 pounds by 4 which is equal to 10 pounds. That is the tial weight of the four books. To find the weight of the fourth book, the three other weights are subtracted from 10 pounds:
10 - 1.7 - 3.2 - 2.8 = 2.3 pounds
FromTheMoon [43]3 years ago
3 0

Answer:

2.3

Step-by-step explanation:

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The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age gro
ra1l [238]

Answer:

a) Null hypothesis:p\leq 0.12  

Alternative hypothesis:p > 0.12  

b) z_{\alpha}=1.64

c) z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362  

d) p_v =P(z>1.362)=0.0866  

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

\hat p=\frac{134}{1000}=0.134 estimated proportion of young drivers ages 18 – 24 that had an accident

p_o=0.12 is the value that we want to verify

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic

p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%:  Null hypothesis:[tex]p\leq 0.12  

Alternative hypothesis:p > 0.12  

The statistic would be given by:

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

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

z_{\alpha}=1.64

Part c

For this case the statistic would be given by:

z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362  

Part d

The p value can be calculated with the following probability:

p_v =P(z>1.362)=0.0866  

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

8 0
3 years ago
Elaine and Rocco shared a bag of popcorn Rocco are 5/8 of the bag Elanie ate 1 3/8.how much popcorn did Elanie and Rocco eat in
amid [387]
2 bags, because when you add 5/8 and 1 3/8 you get 1 8/8 which is really 2
7 0
4 years ago
___ L = 240 kL ? <br><br> A. 24,000<br> B. 240,000<br> C. 0.240<br> D. 2.4
koban [17]
It's b 240000
I'm sorry if wrong I'm not too good at these things
6 0
3 years ago
Read 2 more answers
Hannah has another candle that is 14 cm tall. How fast must it burn in order to also be 6 cm tall after 4 hours? Explain your th
RideAnS [48]

Answer:

The candle must burn at 2cm per hour in order to be 6cm tall after 4 hours. I know this because the candle has to burn 8cm in 4 hours and that amounts to 2cm per hour.

Step-by-step explanation:

Lets solve!

We have to find out how much of the candle needs to be burned!

14cm - 6cm = 8cm

The candle must burn 8cm in 4 hours!

Lets find out how much the candle burns in 1 hour.

\frac{8cm}{4hours}= 2cm per hour

The candle must burn at 2cm per hour in order to be 6cm tall after 4 hours. I know this because the candle has to burn 8cm in 4 hours and that amounts to 2cm per hour.

Woohoo, We did it! <u>Would you like to mark my answer as brainliest?</u> I would love that!

6 0
3 years ago
Subtraction equation with a difference of 54.57
hodyreva [135]
148.57-94 would equal 54.57
5 0
3 years ago
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