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MissTica
3 years ago
9

How do you factor 5x^2b^2 + 14xb^2 -3b^2

Mathematics
1 answer:
Monica [59]3 years ago
3 0

Answer:

5b^2(x-0.2)(x+3)

Step-by-step explanation:

The first thing that I noticed was that all of the terms had a common factor of b^2. You can therefore factor that out first:

5x^2b^2 + 14xb^2 -3b^2= \\\\b^2(5x^2+14x-3)

Now, you have a quadratic equation inside the parentheses. Factoring, you find that the roots are -0.2 and 3, meaning that you can further factor this expression to be:

5b^2(x-0.2)(x+3)

Hope this helps!

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Whats the differnce in (5x+4)-(x+2)
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Answer:

4x+2

Step-by-step explanation:

(5x+4)-(x+2)=(5-1)*x+(4-2)=4x+2

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Create a unique parabola in the pattern f(x) = (x − a)(x − b)
MissTica

Answer:

f(x)=(x-6)(x-(-7))

Step-by-step explanation:

Foiled:

0 = x + x - 42

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3 years ago
Suppose a simple random sample of size 50 is selected from a population with ? = 10. Find the value of the standard error of the
viktelen [127]

Answer:

The population size is N = 50,000

Step-by-step explanation:

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3 years ago
What are these signs for these expressions???
vekshin1

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look at the photo

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If you will need any help,don't hesitate to ask

8 0
4 years ago
Hi, can you help me answer this question please, thank you
BartSMP [9]

3)

Assuming that the population of the study has a normal distribution and that you are studying the population mean μ.

The statistic hypotheses are

\begin{gathered} H_0\colon\mu=87.8 \\ H_1\colon\mu

If you look at the alternative hypotheses, it includes the symbol "<", which indicates that the test is <em>one-tailed</em>. This means that you will reject the null hypotheses at small values of the test statistic.

Since the sample size is small (n=4) and the given sample data. The test statistic to use for the test is the Student's t:

t=\frac{\bar{X}-\mu}{\frac{S}{\sqrt[]{n}}}\sim t_{n-1}

To calculate the value of the statistic under the null hypothesis you have to use the sample data and the value of the population mean under the null hypothesis:

μ= 87.8

sample mean= 75

sample standard deviation= 17.1

n=4

\begin{gathered} t_{H0}=\frac{75-87.8}{\frac{17.1}{\sqrt[]{4}}} \\ t_{H0}=\frac{-12.8}{8.55} \\ t_{H0}=-1.497 \end{gathered}

The test statistic under the null hypothesis is -1.497

The p-value is the probability corresponding to the calculated statistic if possible under the null hypothesis.

So you have to calculate the probability of obtaining the value t=-1497

The student t has n-1 degrees of freedom, since the sample taken is n=4, the degree of freedom is 3.

The p-value is then the accumulated probability up to -1.497 under a t-distribution with 3 degrees of freedom:

p-\text{value}=P(t_3\leq-1497)=0.115653\approx0.1157

The p-value is equal to 0.1157

4)

To make a decision using the p-value, you have to compare the said value with the level of significance, following the decision rule:

- If the p-value ≥ α, do not reject the null hypothesis.

- If the p-value < α, reject the null hypothesis.

The p-value (0.1157) is greater than the significance level (α= 0.05)

The decision is "do not reject the null hypothesis."

5 0
1 year ago
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