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Natali5045456 [20]
3 years ago
10

What is the area of a triangle with a base of 15 1/4 inches and a height of 10 inches?

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Area of the triangle is 76.25 in²

Step-by-step explanation:

  • Step 1: Find the area of the triangle with base = 15 1/4 in and height = 10 in

Area of the triangle = 1/2 × base × height

                                 = 1/2 × 61/4 × 10 = 76.25 in²

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Determine the equation of a line parallel to the graph of y=3x - 9 and goes through the point (2, 10)​
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Step-by-step explanation:

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Answer:

s_p =\sqrt{\frac{(12 -1)(5100)^2 +(12-1)(5900)^2}{12 +12 -2}}=5514.526

t=\frac{(37900-39800)-0}{5514.526\sqrt{\frac{1}{12}+\frac{1}{12}}}}=-0.844  

p_v =2*P(t_{22}  

Comparing the p value with a significance level for example \alpha=0.1 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 conclude that the true means are not significantly different.  

Step-by-step explanation:

Data given and notation

\bar X_{A}=37900 represent the mean for A

\bar X_{B}=39800 represent the mean for B

s_{A}=5110 represent the sample standard deviation for A

s_{B}=5900 represent the sample standard deviation for B

n_{A}=12 sample size for the group A  

n_{B}=12 sample size for the group B

\alpha 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 means are equal or not, the system of  hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{B}= 0  

Alternative hypothesis:\mu_{A} - \mu_{B}\neq 0  

We don't have the population standard deviation's but we assume that the population deviation is equal for both populations, so we can apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{s_p\sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} (1)  

Where s_p represent the standard deviation pooled given by:

s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}

s_p =\sqrt{\frac{(12 -1)(5100)^2 +(12-1)(5900)^2}{12 +12 -2}}=5514.526

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{(37900-39800)-0}{5514.526\sqrt{\frac{1}{12}+\frac{1}{12}}}}=-0.844  

P value  

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

df=n_A +n_{B}-2=12+12-2=22

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

p_v =2*P(t_{22}  

Comparing the p value with a significance level for example \alpha=0.1 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 conclude that the true means are not significantly different.  

6 0
3 years ago
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