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Anna007 [38]
3 years ago
13

John calculated the discriminant of f(x) = 5x² + 3x + 1 to be a negative number (-11). Which of the following is correct? A. the

re are no real roots B. there are 2 real roots C. there is one real root D. there are 2 negative roots
Mathematics
1 answer:
saul85 [17]3 years ago
4 0
The discriminant is a parameter of an object calculated as an aid to its solution. in a quadratic expression, discriminant is given by (b^2-4ac), when the discriminant is a negative number it implies that:
A. There are no real roots
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Answer:

t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604  

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So the p value is a very low value and using any significance level for given \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

\bar X_{1}=9.1 represent the mean for the sample 1

\bar X_{2}=8 represent the mean for the sample 2

s_{1}=1.9 represent the sample standard deviation for the sample 1

s_{2}=2.1 represent the sample standard deviation for the sample 2

n_{1}=40 sample size for the group 1

n_{2}=50 sample size for the group 2

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Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis:\mu_{1} = \mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (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.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604  

Statistical decision

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=40+50-2=88

Since is a bilateral test the p value would be:

p_v =2*P(t_{(88)}>2.604)=0.0108

So the p value is a very low value and using any significance level for given \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

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