Plz answer quickly !!!!

Does anyone know how to do this? Im so confused.

Dafna11 [192]

<h3>Answer: PC = 5</h3>

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Work Shown:

AB+BC+CD = AD .... segment addition postulate

BC+BC+CD = AD ... replace AB with BC (since AB = BC)

BC+BC+BC = AD ... replace CD with BC (since BC = BC)

3*BC = AD .... combine like terms

3*BC = 12 .... replace AD with 12

BC = 12/3 .... divide both sides by 3

BC = 4

For right triangle PBC, the legs are BP = 3 and BC = 4. We can use the Pythagorean Theorem to find that the hypotenuse is PC = 5

The steps are shown below

(BP)^2+(BC)^2 = (PC)^2

3^2+4^2 = (PC)^2

9+16 = (PC)^2

25 = (PC)^2

(PC)^2 = 25

PC = sqrt(25)

PC = 5

The other parts of the diagram seem to be thrown in as a distraction.

7 0

3 years ago

3 + (5 + 4) = 3+ (4 + 5) illustrates which property?

o-na [289]

Answer:

Distributive

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The speed of sound in air is 340 m/s. How far , in meters, would a sound of wave travel on 7 seconds?

Rasek [7]

Answer:

2 380m

Step-by-step explanation:

very simple just multiply 7×340

4 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .