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3 years ago

How do I factor an expression using the GCF?

Mathematics

1 answer:

Anni [7]3 years ago

4 0

Answer:

use the distributive property

Step-by-step explanation:

Use the distributive property to rewrite the expression with the GCF factored out.

Example:

ab +ac +ad . . . . has terms with a common factor of 'a'

When the common factor is factored out, the distributive property tells you the equivalent is ...

= a(b +c +d)

If 'a' is the greatest common factor, then b, c, d are mutually prime and no more factors can be removed to outside the parentheses.

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What is the slope of the line? What is the y-intercept? Write the equation of the line

ohaa [14]

Answer:

slope is 2

y-intercept is 3

equation y=2x+3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted

hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ = 0.

Hₐ: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ ≠ 0.

Assume that the significance level of the test is, α = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

Compute the pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16$

The critical value of t is:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984$

*Use a t-table.

Compute the 95% confidence interval for mean difference as follows:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

$=(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}$

$=-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)$

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0

3 years ago

I really need help on this one

CaHeK987 [17]

He has the same as well 6.5

3 0

3 years ago

25=35x What does x equal?

jeka57 [31]

Answer:

$\frac{25}{35}$ , which simplifies into $\frac{5}{7}$