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4 years ago

Factor 13ab3 + 39a2b5.

Mathematics

2 answers:

lozanna [386]4 years ago

5 0

$13ab^3+39a^2b^5\\\\\boxed{\boxed{\boxed{13ab^3(1+3ab^2)}}}\\\\$

Brazil number one.

borishaifa [10]4 years ago

3 0

Answer:

there's no answer for that equation

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If i drive at a constant rate and go 100 miles in 2 hours, write an equation that could represent the number of miles, m, per ho

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100 miles ............... 2 hours
x miles ................1 hour

x = 100 m / 2 h = 50 m / h

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4 years ago

10 kids are randomly grouped into an A team with five kids and a B team with five kids. Each grouping is equally likely. (a) Wha

nikitadnepr [17]

Answer:

I think B is 1/5. I think C is 1/2.

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3 years ago

Read 2 more answers

12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th

atroni [7]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

$Confidence= 0.95$ the confidence level desired

$ME =0.03$ represent the margin of error desired

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

The confidence level is 95% or 0.95, the significance is $\alpha=0.05$ and the critical value for this case using the normal standard distribution would be $z_{\alpha/2}=1.96$

Since we don't have prior information we can use $\hat p= 0.5$ as an unbiased estimator

Also we know that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

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3 years ago

Simplify (tan(θ)+cot(θ))/(sec(θ)•csc(θ)) completely

AlekseyPX

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3 years ago

8-4 trigonometry (part b) hw (2nd part)

marishachu [46]

Answer:

82+25=107 yan po ang answer

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3 years ago