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Viefleur [7K]
3 years ago
9

Find the gcf of 21 and 30

Mathematics
1 answer:
Vanyuwa [196]3 years ago
8 0

3 is the gcf of 21 and 30


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A proportion is an equation showing the equivalence of two ratios or rates true or false
Lynna [10]
The answer is true- A proportion is an equation showing equivalent of two ratios or rates
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3 years ago
Compute: 5 ∙ (2 + 3i)
dimulka [17.4K]

Answer:

It will be equal to 40

Step-by-step explanation:

We have to compute 5\times (2+3!)

Let first find 3 !

For this we have to use factorial concept

So 3! will be equal to 3! = 3×2×1 = 6

Now According to 6+2 = 8

And now after solving bracket we have to multiply with 5

So 5×8 = 40

So after computation 5\times (2+3!)=40

So the final answer will be 40

4 0
3 years ago
At pretend highschool there are 325 boys and 275 girls.
RideAnS [48]

Answer:

13:11

Step-by-step explanation:

325:275

If divided by 5 all through, you get 65:55, then by 5 again you get 13:11

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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
Duke ate 1.2 kilograms of dog food each day for a total of 8.4 kilograms while his owners were out of town during the month of J
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The awnser is d, March.
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