Please help me out. Don't answer if you don't know the answer. NO LINKS!!

280 divided by 8 equal

IgorC [24]

Answer:

Step-by-step explanation:

8 0

2 years ago

18. Mabel gave Margo a bag containing red, white, and

Sedbober [7]

So if the bag contains half as many red marbles as blue ones, and there are 8 reds then there are 16 blues. 16 + 8= 24 if you want there to be 1/4 white marbles add 6. 6/30 is 1/4.

So the answer comes out to be 30 marbles in the bag

4 0

2 years ago

Read 2 more answers

The average of five numbers is 85. If each of the numbers is increased by 4, what is the average of the five new numbers?

velikii [3]

Answer:

21.25!!!

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e.

zaharov [31]

Question

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:_Question

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:________Question

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:______________QuQuestion

Show that for a square Question Question

Show that for a square symmetric matrix M, Question

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:___________any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:___________Question

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:___________

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:___________tric mQuestion

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:___________atrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:___________estion

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:______________Question

Show that for a square symmetric matrix M, any two eigen-vectors v1, v2 with distinct eigen-values λ1, λ2, are orthogonal, i.e. inner product of v1 and v2 is zero. This shows that a symmetric matrix has orthonormal eigen-vectors:__________________

3 0

3 years ago

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents di

cupoosta [38]

Answer:

The probability value is almost equal to 0. Implying that the proportion of people dying in that particular interval if deaths occurred randomly throughout the year is unusual.

Step-by-step explanation:

The random variable X can be defined as the number of decedents who died in the three-month period following their birthdays.

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents died in the three-month period following their birthdays (123).

The probability (p) of anyone dying in any quarter if people die randomly during the year is simply 0.25.

The random variable X follows a Binomial distribution with parameters n = 747 and p = 0.25.

But the sample selected is too large and the probability of success is small.

So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=747\times 0.25=186.75>10\\n(1-p)=747\times (1-0.46)=560.25>10$

Thus, a Normal approximation to binomial can be applied.

So, $p\sim N(\hat p,\ \frac{\hat p(1-\hat p)}{n})$ .

Compute the probability that 46% or more would die in that particular interval if deaths occurred randomly throughout the year as follows:

$P (p\geq 0.46)=P(\frac{p-\hat p}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}>\frac{0.46-0.25}{\sqrt{\frac{0.25(1-0.25)}{747}}})$

$=P(Z>13.25)\\=1-P(Z$

*Use a z table for the probability.

The probability value is almost equal to 0. This probability is very low indicating that the proportion of people dying in that particular interval if deaths occurred randomly throughout the year is unusual.

3 0

2 years ago