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3 years ago

Can someone help me with this question I’m confused

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

3 0

Answer:

$8,466.15

Step-by-step explanation:

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Please help ASAP 20 points to whoever helps!

IceJOKER [234]

Answer:

A. 232 in²

B. 216 in²

Step-by-step explanation:

A. Box A is a cuboid/rectangular prism.

Surface area of box A = 2(LW + LH + WH)

L = 10 in

W = 2 in

H = 8 in

Surface area = 2(10*2 + 10*8 + 2*8)

= 2(20 + 80 + 16) = 2(116)

Surface area = 232 in²

B. Surface area of B = surface area of cube

Surface area of cube = 6s²

s = 6 in

Surface area = 6(6²)

= 6(36)

= 216 in²

5 0

3 years ago

PLEASE HELP!! (Look at photo)

Mila [183]

Answer:

10 x (p+2) = 30

10 x 2 = 20 30 - 20 = 10 10 ÷ 10 = 1

P = 1

10 x (1 + 2) = 30

5 0

3 years ago

It’s geometry I need help

Gnom [1K]

Answer: Parallel

Y=x will be on the point of the graph
Y= x+2 will be two units up but still positive so they will never interest

6 0

3 years ago

Read 2 more answers

According to national data, 5.1% of burglaries are cleared with arrests. A new detective is assigned to six different burglaries

blagie [28]

Answer:

26.95% probability that at least one of them is cleared with an arrest

Step-by-step explanation:

For each burglary, there are only two possible outcomes. Either it is cleared, or it is not. The probability of a burglary being cleared is independent of other burglaries. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

5.1% of burglaries are cleared with arrests.

This means that $p = 0.051$

A new detective is assigned to six different burglaries.

This means that $n = 6$

What is the probability that at least one of them is cleared with an arrest?

Either none are cleared, or at least one is. The sum of the probabilities of these events is 100% = 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.051)^{0}.(0.949)^{6} = 0.7305$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7305 = 0.2695$

26.95% probability that at least one of them is cleared with an arrest

8 0

3 years ago

The matrix A = −14 −6 6 28 12 −4 0 0 4 has characteristic polynomial

rosijanka [135]

Answer:

Characteristic equation:

$p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda$

Eigen values:

$\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4$

Step-by-step explanation:

We are given the matrix:

$\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]$

The characteristic equation can be calculated as:

$det(A-\lambda I) = 0\\|A-\lambda I| = 0$

We follow the following steps to calculate characteristic equation:

$=det\Bigg(\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\Bigg)\\\\= det\Bigg(\displaystyle\left[\begin{array}{ccc}-14-\lambda&-6&6\\28&12-\lambda&-4\\0&0&4-\lambda\end{array}\right]\Bigg)\\\\=(-14-\lambda)[(12-\lambda)(4-\lambda)]+6[28(4-\lambda)]-6[(28)(0)-(12-\lambda)(0)]\\\\= -\lambda^3 + 2\lambda^2 + 8\lambda$

$p(\lambda)= -\lambda^3 + 2\lambda^2 + 8\lambda$

To obtain the eigen values, we equate the characteristic equation to 0:

$p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda = 0\\-\lambda(\lambda^2-2\lambda-8) = 0\\-\lambda(\lambda^2-4\lambda+2\lambda-8) = 0\\-\lambda[(\lambda(\lambda-4)+2(\lambda-4)] = 0\\-\lambda(\lambda+2)(\lambda-4) = 0 \\\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4$

We can arrange the eigen values as:

$\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4\\-2 < 0 < 4\\\lambda_2 < \lambda_1 < \lambda_3$

3 0

3 years ago