The following conditional statement true. what is the statements converse and is the converse true?

1 answer:

4 0

A converse is when we switch the hypothesis and conclusion, so "if a figure is a square then it has four right angles" the converse would be "If a figure has four right angles then it is a square" Which is false because of rectangles, your answer is D.

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HELP NOW PLEASE!<br>What is the volume of a cube with length 8 feet? <br>V=lwh <br><br> _ft³

Igoryamba

the volume of the cube is: 512 cubic feet

Step-by-step explanation:

Given

Length = l = 8 feet

The formula for the volume of a cube is given by:

$V = l * w * h$

In case of a cube, the length, width and height are equal so the volume will be:

$V = l^3$

Putting l = 8

$V = (8)^3\\V = 8*8*8\\V = 512\ ft^3$

Hence,

the volume of the cube is: 512 cubic feet

Keywords: Volume, cube

Learn more about volume at:

#LearnwithBrainly

5 0

4 years ago

Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,

OLEGan [10]

Let $D$ denote the event that an HD is defective, and $F_i$ the event that a particular HD was produced at facility $i$ .

You are asked to compute

$\mathbb P(F_2\mid D)$
$\mathbb P(F_4\mid D)$
$\mathbb P(D^C)$

From the definition of conditional probabilities, the first two will require that you first find $\mathbb P(D)$ . Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
$\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}$

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so $\mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015$ .

Use the law of total probability to determine the value of the denominator:

$\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)$

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

$\mathbb P(D)=0.015+0.02+0.01+0.03=0.075$

and thus

$\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04$

(b) Similarly,
$\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16$

(c)
$\mathbb P(D^C)=1-\mathbb P(D)=0.925$

6 0

3 years ago

If two cards are randomly chosen from a standard deck, what is the probability that a 6 is chosen, NOT replaced, and another 6 i

Rama09 [41]

$|\Omega|=52\cdot51=2652\\ |A|=4\cdot3=12\\\\ P(A)=\dfrac{12}{2652}\Rightarrow \text{D}$