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joja [24]
2 years ago
11

At a hospital, 56 percent of the babies born are boys. of the baby girls born, 12 percent are premature. what is the probability

of a premature baby girl being born at this hospital? round to the nearest percent. 5% 7% 21% 27%
Mathematics
2 answers:
Phantasy [73]2 years ago
8 0

Answer: First Option (5%)

Step-by-step explanation:

Rudik [331]2 years ago
4 0
If the likelihood of boys being born is 56%, then the complement of that is the likelihood of girls being born which is 100% - 56% = 44%.
Since we are given that of the baby girls born, 12% are premature, the probability of a premature baby girl  being born is:
(0.44)(0.12) = 0.0528 or 5.28%
The closest answer is the FIRST OPTION (5%).
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Ethan bought 4 packages of pencils. After he gave 8 pencils to his friends, he had 40 pencils left over. How many pencils were i
swat32

Answer:

12 pencils in each package.

Step-by-step explanation:

Given:

Ethan bought 4 packages of pencils.

No of Packages = 4

No of pencils given to friend = 8 pencils

No of pencils left over =40 pencils

∴Total Number of Pencils = No of pencils given to friend + No of pencils left over =40+8 =48

Total Number of pencils in each package = \frac{\textrm{Total Number of Pencils}}{\textrm{No Of Packages}}= \frac{48}{4}= 12 \ pencils

8 0
2 years ago
A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st
maxonik [38]

Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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The diagram included should explain this.

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Same 10 tens I guess
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