What is 378 times by 9

Circle K has a diameter of 10 centimeters and a major arc RS is intercepted by a central angle of 190°. Find the length of major

drek231 [11]

Answer:

16.6

Step-by-step explanation:

7 0

3 years ago

Given that the product of 22 integers is equal to 1, prove that their sum is not equal to 0.

Kay [80]

1. If the product of these integers is to be 1, then all of them must be either 1 or -1.

2. Since the product is positive (+1), it must be that there are an *even* number of negative ones (-1), if any.

3. If the sum were 0 it would mean that the number of +1's must equal the number of -1's. So that means there would have to be exactly 22/2=11 of each.

4. But if there were 11 of each, that means the number of -1's would be *odd* and there's no way the product could be +1 (as stated in 2 above).

Hence, the sum is never 0, if the product of 22 integers is equal +1.

4 0

3 years ago

Read 2 more answers

What is (-3,5) reflected across the y axis?

BARSIC [14]

3,5 is the answer i belive

8 0

3 years ago

M= 2²×3×5² N=2³×3²×7 find the largest Number that divides exactly into M and N

nevsk [136]

Answer:

12

Step-by-step explanation:

just find the HCF, which will be

2² × 3 = 12

5 0

2 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago