Dwayne bought 3 yards of wrapping paper how many inches is that?

Write 9− 16x⁴ as the product of two factors.

Marysya12 [62]

Answer:

The two factors will be $=(3+4x^2)\ and\ (3-4x^2)$

Step-by-step explanation:

We have given the equation $9-16x^4$

We know the algebraic equation $a^2-b^2=(a+b)(a-b)$

As $9-16x^4$ can be written as $3^2-(4x^2)^2$ , as square of 3 is 9 and square of $4x^2$ is $16x^4$

So $9-16x^4=3^2-(4x^2)^=(3+4x^2)(3-4x^2)$

So the two factors will be $=(3+4x^2)\ and\ (3-4x^2)$

6 0

3 years ago

If a4 = 625, what does (a) equal?

Reika [66]

I'm assuming you meant to write a^4 = 625.

If that is the case, then note how 625 = 25^2, and how a^4 is the same as (a^2)^2

So we go from this

a^4 = 625

to this

(a^2)^2 = 25^2

Apply the square root to both sides and you'll end up with: a^2 = 25

From here, apply the square root again to end up with the final answer: a = 5 or a = -5

As a check:

a^4 = (-5)^4 = (-5)*(-5)*(-5)*(-5) = 25*25 = 625

a^4 = (5)^4 = (5)*(5)*(5)*(5) = 25*25 = 625

Both values of 'a' work out

4 0

3 years ago

What is the solution of log(3x − 2)125 = 3?

Reil [10]

$log_{3x - 2}125 = 3 \\ (3x - 2)^3 = 125 \\ \sqrt[3]{(3x - 2)^3} = \sqrt[3]{125} \\ 3x - 2 = 5 \\ 3x = 5 + 2 \\ 3x = 7 \\ x = 7/3$

8 0

3 years ago

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi

boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Description in words of the parameter p

$p$ represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that have experienced bullying

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula

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

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.