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Len [333]
3 years ago
12

What is 285 - 15? 17 28 19 21 Please help quickly thanks

Mathematics
1 answer:
kifflom [539]3 years ago
7 0

Answer:

19

Step-by-step explanation:

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How do i solve this problem 1/15+1/6
Semenov [28]

Answer:

7/30

Step-by-step explanation:

1/6 = 1x5/6x5

    = 5/30

1/15= = 1x2/1x15

       =2/30

2/30 + 5/30 = 7/30

5 0
3 years ago
Can someone help me with this?
tino4ka555 [31]

Answer:

just add both sides 2.8+2.8=5.6

Step-by-step explanation: possible is 5.6 just put 2.8 on both

7 0
3 years ago
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How do u write 32/100 in expanded form
laiz [17]
In expanded form or decimal form 32/100 is equal to 0.32 or 32% or if simplified the fraction is 8/25
3 0
3 years ago
Read 2 more answers
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
How many ways are there to choose a half dozen donuts from 10 varieties a)If there are no two donuts of the same variety.b)If th
777dan777 [17]

Answer:

If multiples are allowed (like "10 plain, two chocolate"), then for each of 12 donuts there are 21 possibilities.

That is 21 x 21 x 21 .... x 21 (12 terms in all) = 21^12

Step-by-step explanation:

7 0
3 years ago
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