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3 years ago

Given the figure below, find the values of x and z. 29 (8x + 37) • (10x - 1)

Mathematics

1 answer:

pantera1 [17]3 years ago

7 0

Answer:

idk

Step-by-step explanation:

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Math slope helpppp meeee

Ganezh [65]

Answer:

1. -3/8 2. 1/2 3. -3/2

Step-by-step explanation:

You can use Y2-Y1/X2-X1 to find the answers

Ask me if you want it more in detail

6 0

3 years ago

If your chances of losing the shell game if you randomly pick is 2 in 3. What are the chances that you would lose 5 games in a r

NISA [10]

Answer:

Step-by-step explanation:

4 0

3 years ago

Write and simplify the integral that gives the arc length of the following curve on the given interval b. If necessary, use tech

LiRa [457]

Answer:

L = 4.103

Step-by-step explanation:

we have length of curve

$L = \int\limits^b_a {\sqrt{(f'(x))²+1} } \, dx$

where $f(x) = d/dx(3*in(x)) = 3/x$

substituting for f(x), we have $L = \int\limits^5_2 {\sqrt{(3/x)²+1} } \, dx$

(since the limit is 2≤ x ≤5)

solving, $L = \int\limits^5_2 {\sqrt{9/x²+1} } \, dx$

Simplifying this integral, we have

L = 4.10321

8 0

3 years ago

To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo

mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

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

Given:

$\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25$

Confidence level = 80%

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397$

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

$CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)$

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0

3 years ago

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago