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lesya [120]
3 years ago
13

Please explain how to solve this problem (in picture)

Mathematics
2 answers:
Archy [21]3 years ago
8 0
Hopefully this helps sorry

Harlamova29_29 [7]3 years ago
5 0
Then the answer is above

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A single die is rolled. find the probability of rolling a 2 or an odd number
prisoha [69]
The chances are 4/6, which can be reduced to 2/3
3 0
3 years ago
Help, please the question and thank you
Deffense [45]

Answer:

Whole number

Step-by-step explanation:

That's only because it is just 1 number that is gonna be there and i did some research too

4 0
3 years ago
Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.
Kobotan [32]

Answer:

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

Step-by-step explanation:

Data given and notation

\bar X_{A}=3.7 represent the mean for athletes  

\bar X_{NA}=4.1 represent the mean for non athletes  

s_{A}=1.1 represent the sample standard deviation for athletes  

s_{NA}=1.3 represent the sample standard deviation for non athletes

n_{A}=32 sample size for the group 2  

n_{NA}=45 sample size for the group 2  

\alpha=0.01 Significance level provided  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the population mean for athletes is lower than the population mean for non athletes, the system of  hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{NA}\geq 0  

Alternative hypothesis:\mu_{A} - \mu_{NA}< 0  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{A}-\bar X_{NA})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{NA}}{n_{NA}}}} (1)  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

P value  

We need to find first the degrees of freedom given by:

df=n_A +n_{NA}-2=32+45-2=75

Since is a one left tailed test the p value would be:  

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

5 0
3 years ago
Whatisthevalueof5.8dividedby1.2
Oliga [24]
 5.8 / 1.2 = 4.83 I hope this is helpful!
3 0
3 years ago
If 42% of a number N is 8.4 what is the value of N
user100 [1]

Answer:

N = 20

Step-by-step explanation:

Symbolically, 0.42N = 8.4.

Solve for N by dividing both sides by 0.42:

         8.4

N = ----------- = 20

        0.42

3 0
3 years ago
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