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3 years ago

How do you change the mix number 66 2/3 into a regular number?

Mathematics

1 answer:

gavmur [86]3 years ago

6 0

Denominator times whole number plus numerator, so:
66x2+3=135

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Can someone help me plead

andrew-mc [135]

Answer:

a. 2^3

b. 3^4

c. 4^3 × 5^2

d. 9^4 × 7^2

Step-by-step explanation:

The following equations are given

a. 2 × 2× 2

b. 3 × 3 × 3 × 3

c. 4 × 4 × 4 × 5 × 5

d. 9 × 7 × 9 × 9 × 7 × 9

We need to find the index notation for the above equations

a. 2^3

b. 3^4

c. 4^3 × 5^2

d. 9^4 × 7^2

In this way it should be done

The same would be relevant

4 0

2 years ago

Brainly which is the distance between (-3,4) and (7,13)

Veseljchak [2.6K]

Answer:

Root 181

Step-by-step explanation:

Using distance formula, we know that the answer will be the square root of 181. Unfortunately, this is not a perfect square, and it is prime, so that number will remain the same.

Distance Formula: $\sqrt{(13-4)^{2} + (7+3})^2} = \sqrt{181}$

Hope this helped!

5 0

3 years ago

Suppose that a box of 86 circuits is sent to a manufacturing plant. Of the 86 circuits shipped, 5 are defective. The plant manag

Triss [41]

Answer:

the probability that the shipment is accepted is 0.8865

Step-by-step explanation:

Given the data in the question;

N = 86, n/d = 5 and n = 2

now, without replacement

the probability that the shipment is accepted will be;

probability that the shipment is accepted = probability that non is defectives

so p( non is defective ) = ( (86-5)/86) × ((86-5-1)/(86-1))

p( non is defective ) = ( 81 / 86) × (80/85)

p( non is defective ) = 0.8865

Therefore, the probability that the shipment is accepted is 0.8865

7 0

2 years ago

Neeed math hellppp!!!!

aleksklad [387]

Answer: A

Step by step: None

6 0

3 years ago

Read 2 more answers

When playing the roulette at a casino, you have exactly 1/35 chance of winning in each round if you play a single number. Your f

Vikentia [17]

Answer:

0.2275 = 22.75% probability that you actually won that round

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Fireworks going off

Event B: You won

Probability of fireworks going off.

100% of 1/35 = 0.0286(when you win)

10% of 34/35 = 0.9714(you lost). So

$P(A) = 0.0286 + 0.1*0.9714 = 0.12574$

Probability of you winning and fireworks going off:

100% of 1/35, so $P(A \cap B) = 0.0286$

If you failed to see the outcome of a round, but you see the fireworks going off, then what is the probability that you actually won that round?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0286}{0.12574} = 0.2275$

0.2275 = 22.75% probability that you actually won that round

8 0

3 years ago