Miller Nolan

What is the measure of ZP, to the nearest degree? O 44° O 46° 0 58° 0 72°

sasho [114]

Answer:

44

Step-by-step explanation:

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6 0

3 years ago

Help me on this Geometry question "Finding angle measures using triangles".

vladimir1956 [14]

X is 63 degrees!
It is equal to the angle marked 63 degrees (angles in that arrangement are always equal)

5 0

3 years ago

Read 2 more answers

Find the Volume. Use pi=3.14. Round to the nearest hundredth.

aleksandr82 [10.1K]

<h2>Volume = 4.19 inches³</h2>

Step-by-step explanation:

radius = 1 inches

π = 3.14

volume of SPHERE = 4/3 × π × radius³

= 4/3 × 3.14 × 1³

= 4.19 inches³

4 0

3 years ago

P(a)=0.50 p(b)=0.30 and p(a and b)=0.15 what is p(a or b)

Softa [21]

Answer:

Option C is correct

P(A or B) = 0.65

Step-by-step explanation:

Given:

P(A) =0.5

P(B)=0.30

P(A and B) =0.15

( The probability of the happening of both independent events will be there product) P( A and B ) =P(A).P(B)

Solution:

To find the probability of the Happening of event A either event B we will use the following formula

P(A or B) = P(A) + P(B)-P(A and B)

= 0.5 + 0.3 - 0.15

=0.65

6 0

3 years ago

Read 2 more answers

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