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3 years ago

Samantha is making brownies for the entire staff at Arlington Elementary School

1 answer:

7 0

It’s 5 boxes because..

4 x 16 = 64 so that wouldn’t be enough

However:

5 x 16 = 80 which is enough for 73 people

Hope this helps! Have a nice day :)

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1. Solve the following simultaneous equations using the matrix method.

dmitriy555 [2]

(i) Use the formula for the determinant of a 2×2 matrix.

$\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad-bc$

$\implies \det(A) = \begin{vmatrix}4 & -3 \\ 2 & 5\end{vmatrix} = 4\times5 - (-3)\times2 = \boxed{26}$

(ii) The adjugate matrix is the transpose of the cofactor matrix of A. (These days, the "adjoint" of a matrix X is more commonly used to refer to the conjugate transpose of X, which is not the same.)

The cofactor of the (i, j)-th entry of A is the determinant of the matrix you get after deleting the i-th row and j-th column of A, multiplied by $(-1)^{i+j}$ . If C is the cofactor matrix of A, then

$C = \begin{pmatrix}5&-2\\3&4\end{pmatrix}$

Then the adjugate of A is the transpose of C,

$\mathrm{adj}(A) = C^\top = \boxed{\begin{pmatrix}5&3\\-2&4\end{pmatrix}}$

(iii) The inverse of A is equal to 1/det(A) times the adjugate:

$A^{-1} = \dfrac1{\det(A)} \mathrm{adj}(A) = \boxed{\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix}}$

(iv) The system of equations translates to the matrix equation

$A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\16\end{pmatrix}$

Multiplying both sides on the left by the inverse of A gives

$A^{-1}\left(A\begin{pmatrix}x\\y\end{pmatrix}\right)=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}$

$\left(A^{-1}A\right)\begin{pmatrix}x\\y\end{pmatrix}=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix} \begin{pmatrix}6\\16\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\boxed{\begin{pmatrix}3\\2\end{pmatrix}}$

4 0

2 years ago

Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value u

musickatia [10]

From the z table the critical value corresponding to the level of significance level for a two-tailed test is z = 1.64.

<h3>How do form the hypotheses?</h3>

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test.

The hypothesis which we put against the null hypothesis is an alternate hypothesis.

The null hypothesis is the one that researchers try to disprove.

From the question, we are told that

The number of observations is greater than 50 i.e. n > 50

The level of significance is $\alpha = 0.1$

Generally given that n > 30 then we will make use of the z table

So From the z table the critical value corresponding to the level of significance level for a two-tailed test is z = 1.64.

Learn more about null and alternative hypotheses here:

brainly.com/question/18831983

#SPJ4

7 0

3 years ago

Which statement describes the sequence defined by a Subscript n Baseline = StartFraction n cubed minus n Over n squared + 5 n En

Dominik [7]

Answer:

The answer is " The sequence converges to infinity. "

Step-by-step explanation:

Given:

$\to a_n=\frac{n^3-n}{n^2+5n}$

$\lim_{n \to \infty} a_n= \lim_{n \to \infty} \frac{n^3-n}{n^2+5n}$

$= \lim_{n \to \infty} \frac{n(n^2-1)}{n(n+5)}\\\\= \lim_{n \to \infty} \frac{(n^2-1)}{(n+5)}\\\\= \lim_{n \to \infty} \frac{n(n-\frac{1}{n})}{n(1+\frac{5}{n})}\\\\= \lim_{n \to \infty} \frac{(n-\frac{1}{n})}{(1+\frac{5}{n})}\\\\$