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3 years ago

Dan buys a car for £3200.

Mathematics

1 answer:

Nataly [62]3 years ago

4 0

Answer:

£2509

Step-by-step explanation:

3200x0.941*4 =2509

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Expand (x – 4)^5 using the Binomial Theorem and Pascal’s triangle

worty [1.4K]

$5th \: row = 1 \: \: \: 5 \: \: \: 10 \: \: \: 10 \: \: \: 5 \: \: \: 1$

$(x - 4) {}^{5} = x {}^{5} + 5x {}^{4} ( - 4) {}^{1} + 10x {}^{3} ( - 4) {}^{2} + 10x {}^{2} ( - 4) {}^{3} + 5x( - 4) {}^{4} + ( - 4) {}^{5}$

$(x - 4) {}^{5} = x {}^{5} - 20x {}^{4} + 160x {}^{3} - 640x {}^{2} + 1280x - 1024$

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2 years ago

Solve the following system of linear equations: 3x1+6x2+6x3 = -9 -2x1–3x2-3x3 = 3 If the system has infinitely many solutions, y

choli [55]

Answer:

The set of solutions is $\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}$

Step-by-step explanation:

The augmented matrix of the system is $\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]$ .

We will use rows operations for find the echelon form of the matrix.

In row 2 we subtract $\frac{2}{3}$ from row 1. (R2- 2/3R1) and we obtain the matrix $\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]$
We multiply the row 1 by $\frac{1}{3}$ .

Now we solve for the unknown variables:

$x_2+x_3=-7$ , $x_2=-7-x_3$
$x_1+2x_2+2x_3=-2$ , $x_1+2(-7-x_3)+2x_3=-2$ then $x_1=12$

The system has a free variable, the the system has infinite solutions and the set of solutions is $\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}$

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3 years ago

The standard deviation for actuary salaries has a standard deviation of $36,730 . You collect a simple random sample of n=36 sal

Yanka [14]

Using the z-distribution, it is found that the lower limit of the 95% confidence interval is of $99,002.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.