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2 years ago

4.3 Rounding to the nearest whole number

Mathematics

2 answers:

Roman55 [17]2 years ago

7 0

1.7=2
8.1=8
6.9=7
4.3=4
Hope this helps

Elan Coil [88]2 years ago

5 0

Answer:

4 is the nearest whole number of 4.3

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Which is the most likely first step needed to solve this equation?

Vanyuwa [196]

Answer:

C. Add 15 to both sides

Step-by-step explanation:

Given equation:

99x – 15 = 45

To solve:

<u>Step 1</u>: Add 15 to both sides:

⇒ 99x – 15 + 15 = 45 + 15

⇒ 99x = 60

<u>Step 2</u>: Divide both sides by 99:

⇒ 99x ÷ 99 = 60 ÷ 99

⇒ x = 60/99

⇒ x = 20/33

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1 year ago

Read 2 more answers

Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win

Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

$P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

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2 years ago

This scatter plot show the relationship between the temperature and amount of sales at a store. The line of best fit is shown on

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Answer:

a) b ≈ 100

b) m ≈ 30

Step-by-step explanation:

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2 years ago

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So: