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timofeeve [1]
3 years ago
12

Which expression is equivalent to 150x² - 60x ?

Mathematics
2 answers:
Bas_tet [7]3 years ago
8 0

Answer: The Answer is B

Step-by-step explanation:

You need to distribute 15x(10x - 4). 15x times 10x is 150x and since there is 2 x's your gonna square it. So then it would be 150x². next 15x times -4 is -60 and add your x. Your answer is 150x² - 60x.

IRISSAK [1]3 years ago
8 0
The answer is B - 100% right
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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  

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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.  

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