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2 years ago

12

Name the intersection of the planes ACE and CDF.

1 answer:

Lynna [10]2 years ago

7 0

Based in your question where as the figure shows the intersection of the planes ACE and CDF and the ask is to name that intersections.Base on my calculation and through step by step procedure and with the guidance of the theories and by the use of formulas, I came up with an answer of the intersection of the two planes is the line CE<span>
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The radius of circle A is 4 times greater than the radius of circle B. Which of the following statements is true?

Black_prince [1.1K]

Answer:

Correct answer: C. Ca = 4 Cb

Step-by-step explanation:

Given: Ra = 4 Rb

Ca : Cb = ?

Using the following formula to calculate the circumference of a circle:

C = 2 R π

Cb = 2 Rb π

Ca = 2 Ra π = 2 ·( 4 · Rb) · π = 4 · (2 Rb π) = 4 · Cb

Ca = 4 · Cb

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3 years ago

What is 70/6 times 18/4 in simplest form

Alenkasestr [34]

70/6 × 18/4

70 * 18/6*4. cross out 18 and 6.

= 70*3/4

= 35 * 2 * 3/ 2*2

= 35*3/2

= 105/2

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3 years ago

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What does P(B/A) = 0.50 mean in terms of the Venn diagram?

Serggg [28]

Answer:

of the 4 people that have a sister, 2 of them have a brother

Step-by-step explanation:

probability of B given A

given that the person has a sister, what is the probability they have a brother

P (B/A) = 1/2

with this venn diagram

of the 4 people that have a sister, 2 of them have a brother

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3 years ago

In a newspaper poll concerning violence on television, 589 people were asked, "What is your opinion of the amount of violence on

abruzzese [7]

Answer:

(a) $P(Y'|M)\approx 0.3297$

(b) $P(Y|M')\approx 0.8323$

(c) $P(Y'|M')\approx 0.1323$

Step-by-step explanation:

Given table is

Yes No Don't Know Total

Men 162 92 25 279

Women 258 41 11 310

Total 420 133 36 589

According the the conditional probability, if A and B are two event then

$P(A|B)=P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

We need to find the following probabilities.

Let Y is the event "saying yes," and M is the event "being a man."

(a)

$P(Y'|M)=\frac{P(Y'\cap M)}{P(M)}$

$P(Y'|M)=\frac{\frac{92}{589}}{\frac{279}{589}}$

$P(Y'|M)=\frac{92}{279}$

$P(Y'|M)=0.329749103943$

$P(Y'|M)\approx 0.3297$

(b)

$P(Y|M')=\frac{P(Y\cap M')}{P(M')}$

$P(Y|M')=\frac{\frac{258}{589}}{\frac{310}{589}}$

$P(Y|M')=\frac{258}{310}$

$P(Y|M')=0.832258064516$

$P(Y|M')\approx 0.8323$

(c)

$P(Y'|M')=\frac{P(Y'\cap M')}{P(M')}$

$P(Y'|M')=\frac{\frac{41}{589}}{\frac{310}{589}}$

$P(Y'|M')=\frac{41}{310}$

$P(Y'|M')=0.132258064516$

$P(Y'|M')\approx 0.1323$

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2 years ago

Easy question please help

iris [78.8K]

Answer:

$17/ $25/$41

Step-by-step explanation:

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2 years ago

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