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3 years ago

8

Rafael was asked to construct a square inscribed in a circle. He drew a circle and a diameter of the circle. Describe how to com

plete his cinstruction. Then, complete the construction.

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

The diameter of the circle must be equal to the diagonal of the square.

Step-by-step explanation:

In order to draw a square inscribed with in the circle, the diameter of the circle must be equal to the diagonal of the square.

a) Draw a circle with center O

b) Draw the diameter of the circle with end points A and B

c) Draw perpendicular bisector of diameter AB intersecting the circle at point C and D

d) Now connect point A, B, C and D

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The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There

tankabanditka [31]

Answer:

The lifetime value needed is 11.8225 hours.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that $\mu = 11, \sigma = 1$ .

What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

This is the value of THE MEAN SAMPLE X when Z has a pvalue of 0.95. That is between Z = 1.64 and Z = 1.65. So we use $Z = 1.645$

Since we need the mean sample, we need to find the standard deviation of the sample, that is:

$s = \frac{\sigma}{\sqrt{4}} = 0.5$

So:

$Z = \frac{X - \mu}{s}$

$1.645 = \frac{X - 11}{0.5}$

$X - 11 = 0.5*1.645$

$X = 11.8225$

The lifetime value needed is 11.8225 hours.

3 0

3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

4 years ago

Suppose a company has fixed costs of $5000 and variable costs per unit of 3 4 x + 1350 dollars, where x is the total number of u

liberstina [14]

Answer:

.

Step-by-step explanation:

3 0

3 years ago

Write a numerical expression for the calculation.

77julia77 [94]

Answer:246.475

Step-by-step explanation:

79+167+19/40

8 0

3 years ago

Find the greatest common factor of -30x4yz3 and 75x4z2.

Papessa [141]

30×4=120 then multiply 75×4=340 then add 120+340=460

3 0

3 years ago