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3 years ago

Can I get help on this please

Mathematics

1 answer:

USPshnik [31]3 years ago

6 0

1. dismal 2. wager 3. peril 4. recline 5. shriek 6. sinister 7. conceal 8. inhabit 9. frigid 10. numb 11. corpse 12. tempt

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A sample of students from an introductory psychology class were polled regarding the number of hours they spent studying for the

Anni [7]

Answer:

$8.68,13.16$

Step-by-step explanation:

Hint- First we have to calculate the mean and standard deviation of the sample and then applying formula for confidence interval we can get the values.

Mean of the sample is,

$\mu=\dfrac{\sum _{i=1}^{24}a_i}{24}=\dfrac{262}{24}=10.92$

Standard deviation of the sample is,

$\sigma =\sqrt{\dfrac{\sum _{i=1}^{24}\left(x_i-10.92\right)^2}{24-1}}=5.6$

The confidence interval will be,

$=\mu \pm Z\dfrac{\sigma}{\sqrt{n}}$

Here,

Z for 95% confidence interval is 1.96, and n is sample size which is 24.

Putting the values,

$=10.92 \pm 1.96\cdot \dfrac{5.6}{\sqrt{24}}$

$=10.92 \pm 2.24$

$=8.68,13.16$

Confidence interval is used to express the degree of uncertainty associated with a sample.

95% confidence interval means that if we used the same sampling method to select different samples and calculate an interval, we would expect the true population parameter to fall within the interval for 95% of the time.

5 0

3 years ago

How to do the inverse of a 3x3 matrix gaussian elimination.

nata0808 [166]

As an example, let's invert the matrix

$\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}$

We construct the augmented matrix,

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]$

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]$

• Multiply row 3 by 1/3 :

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Multiply row 2 by 1/7 :

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

$\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Multiply row 1 by -1/3 :

$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

So, the inverse of our matrix is

$\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}$

6 0

2 years ago

Stephanie found a sweater on sale for $12.88. The original price of the sweater was $44.95. About how much did she save on the s

rusak2 [61]

Answer:

$32.07

ROUNDING ANSWER : $32

Step-by-step explanation:

you would subtract $44.95 -$12.88 which would give you $32.07

since you are rounding your answer would be $32

Hope this helps

8 0

3 years ago

The sum of 10 times a number and fifteen is added to 11 times the same number. (how do i make this into a algebraic expression?)

Ne4ueva [31]

10a times (15+11) or 10a times 26

6 0

3 years ago

Paws at play made a total of $1,234 grooming 22 dogs. Paws at play charges $43 to groom each small dog and $75 for each large do

Nesterboy [21]

Answer:

The system of equation can be used for determine the number of small and the large dogs groomed are

x + y = 22

43x + 75y = 1234

Step-by-step explanation:

Let us assume that the number of small dogs groomed be x.

Let us assume that the number of lagre dogs groomed be y.

As given

Paws at play made a total of $1,234 grooming 22 dogs.

Than the equation becomes

x + y = 22

Paws at play charges $43 to groom each small dog and $75 for each large dog.

Than the equation becomes

43x + 75y = 1234

Therefore the system of equation can be used for determine the number of small and the large dogs groomed are

x + y = 22

43x + 75y = 1234

4 0

3 years ago