The value of expanding (2x -3)^4 is 16x^4 + 96x^3 +216x^2 -216x + 81
<h3>How to expand the expression?</h3>
The expression is given as:
(2x -3)^4
Using the binomial expansion, we have:
Evaluate the combination factors.
So, we have:
Evaluate the exponents and the products
Hence, the value of expanding (2x -3)^4 is 16x^4 + 96x^3 +216x^2 -216x + 81
Read more about binomial expansions at:
brainly.com/question/13602562
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To solve for the confidence interval for the population
mean mu, we can use the formula:
Confidence interval = x ± z * s / sqrt (n)
where x is the sample mean, s is the standard deviation,
and n is the sample size
At 95% confidence level, the value of z is equivalent to:
z = 1.96
Therefore substituting the given values into the
equation:
Confidence interval = 3 ± 1.96 * 5.8 / sqrt (51)
Confidence interval = 3 ± 1.59
Confidence interval = 1.41, 4.59
Therefore the population mean mu has an approximate range
or confidence interval from 1.41 kg to 4.59 kg.
Answer:
Option D. A student who studies 0 hours can expect to score about a 60 on the test
Step-by-step explanation:
we know that
The y-intercept is the value of the y-coordinate when the value of the x-coordinate is equal to zero
In this problem
The x-coordinate represent the hours spent studying
The y-coordinate represent the Test score
so
For x=0, y=60
That means
A student who studies 0 hours can expect to score about a 60 on the test
![\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y = 4\frac{2}{3}x\qquad \qquad yes\qquad \checkmark\qquad \qquad k = 4\frac{2}{3} \\\\[-0.35em] ~\dotfill\\\\ y=3(x-1)\implies \stackrel{\textit{distributing}}{y=3x-3}\qquad \qquad yes\qquad \checkmark \qquad \qquad k=3](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bdirect%20proportional%20variation%7D%20%5C%5C%5C%5C%20%5Ctextit%7B%5Cunderline%7By%7D%20varies%20directly%20with%20%5Cunderline%7Bx%7D%7D%5Cqquad%20%5Cqquad%20y%3Dkx%5Cimpliedby%20%5Cbegin%7Barray%7D%7Bllll%7D%20k%3Dconstant%5C%20of%5C%5C%20%5Cqquad%20variation%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20y%20%3D%204%5Cfrac%7B2%7D%7B3%7Dx%5Cqquad%20%5Cqquad%20yes%5Cqquad%20%5Ccheckmark%5Cqquad%20%5Cqquad%20k%20%3D%204%5Cfrac%7B2%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20y%3D3%28x-1%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bdistributing%7D%7D%7By%3D3x-3%7D%5Cqquad%20%5Cqquad%20yes%5Cqquad%20%5Ccheckmark%20%5Cqquad%20%5Cqquad%20k%3D3)
bear in mind that, direct proportional equations have a y-intercept.
for y = kx, is pretty much y = kx + 0, where 0 = y-intercept.
and the "k" constant of proportionality, is pretty much just its slope.
Decimal : 1.5
Percent : 150%
