Solve the following equation 2x <br> 2<br> −3=0<br> 1

Help please i need this done soon!! tysm!!

olga nikolaevna [1]

Answer:

70

Step-by-step explanation:

Opposite Angles are supplementary in cyclic quadratlerials.

3 0

3 years ago

In the figure, x ∥ y and a is a transversal that crosses the parallel lines.

Dominik [7]

Here, exterior angles are 1, 2, 7 & 8 as they are outside the parallel lines, & among them, alternates are: 1 with 7 and 2 & 8. 1 & 7 is not in option but 2 and 8 is there in C

In short, Your Answer would be Option C

Hope this helps!

8 0

3 years ago

3( 3v +5 ) = 39<br> Solve this problem

V125BC [204]

Answer:

V= 8/3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Consider the following scenario: Let P(C) = 0.4. Let P(D) = 0.5. Let P(C|D) = 0.6. a. Find P(C AND D). b. Are C and D mutually e

Nataly_w [17]

a. By definition of conditional probability,

P(C | D) = P(C and D) / P(D) ==> P(C and D) = 0.3

b. C and D are mutually exclusive if P(C and D) = 0, but this is clearly not the case, so no.

c. C and D are independent if P(C and D) = P(C) P(D). But P(C) P(D) = 0.2 ≠ 0.3, so no.

d. Using the inclusion/exclusion principle, we have

P(C or D) = P(C) + P(D) - P(C and D) ==> P(C or D) = 0.6

e. Using the definition of conditional probability again, we have

P(D | C) = P(C and D) / P(C) ==> P(D | C) = 0.75

3 0

3 years ago

App 19.study

lions [1.4K]

Answer:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

Step-by-step explanation:

Given

$n = 125$

See attachment for proper table

Required

Complete the table

Experimental probability is calculated as:

$Pr = \frac{Frequency}{n}$

We use the above formula when the frequency is known.

For result of roll 2, 4 and 6

The frequencies are 13, 29 and 6, respectively

So, we have:

$Pr(2) = \frac{13}{125} = 0.104$

$Pr(4) = \frac{29}{125} = 0.232$

$Pr(6) = \frac{6}{125} = 0.048$

When the frequency is to be calculated, we use:

$Pr = \frac{Frequency}{n}$

$Frequency = n * Pr$

For result of roll 3 and 5

The probabilities are 0.144 and 0.296, respectively

So, we have:

$Frequency(3) = 125 * 0.144 = 18$

$Frequency(5) = 125 * 0.296 = 37$

For roll of 1 where the frequency and the probability are not known, we use:

$Total \ Frequency = 125$

So:

Frequency(1) added to others must equal 125

This gives:

$Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125$

$Frequency(1) + 103 = 125$

Collect like terms

$Frequency(1) =- 103 + 125$

$Frequency(1) =22$

The probability is then calculated as:

$Pr(1) = \frac{22}{125}$

$Pr(1) = 0.176$

So, the complete table is:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

5 0

3 years ago

Solve the following equation 2x 2 −3=0 1