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2 years ago

Round 0.4283 to the nearest thousandth

Mathematics

1 answer:

FrozenT [24]2 years ago

3 0

The thousandths place is the 3rd number to the right after the decimal, so it would be 0.428

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If m∠a=47°, what is m∠b, m∠c, and m∠d? Explain how you know the measures are correct.

avanturin [10]

m ∠b = 133°, m ∠c = 47°, and m ∠d = 133°.

<h3>Further explanation</h3>

Follow the attached picture. I sincerely hope that's precisely a correct illustration.

We will use a graph of two intersecting straight lines.

Note that m ∠a and m ∠c are vertical angles. Since vertical angles share the same measures, in other words always congruent, we see $\boxed{ \ m \ \angle{c} = m \ \angle{a} \ } \rightarrow \boxed{\boxed{ \ m \ \angle{c} = 47^0 \ }}$

We continue to determine m ∠b and m ∠d.

Note that m ∠b and m ∠d represent supplementary angles. Recall that supplementary angles add up to 180°.

Let us see the following steps.

$\boxed{ \ m \ \angle{a} + m \ \angle{b} = 180^0. \ }$

$\boxed{ \ m \ 47^0 + m \ \angle{b} = 180^0. \ }$

Both sides subtracted by 47°.

$\boxed{ \ m \ \angle{b} = 180^0 - 47^0. \ }$

Thus $\boxed{\boxed{ \ m \ \angle{b} = 133^0. \ }}$

Finally, note that m ∠b and m ∠d are vertical angles. Accordingly, $\boxed{ \ m \ \angle{d} = m \ \angle{b} \ } \rightarrow \boxed{\boxed{ \ m \ \angle{d} = 133^0 \ }}$

<u>Conclusion:</u>

m ∠a = 47°
m ∠b = 133°
m ∠c = 47°
m ∠d = 133°

<u>Notes:</u>

Supplementary angles are two angles when they add up to 180°. $\boxed{ \ example: \angle{a} + \angle{b} = 180^0 \ }$
Vertical angles are the angles opposite each other when two lines cross. Note that vertical angles are always congruent, or of equal measure. $\boxed{ \ example: \angle{a} = \angle{c} \ }$

<h3>Learn more</h3>

About the measure of the central angle brainly.com/question/2115496
Undefined terms needed to define angles brainly.com/question/3717797
Find out the measures of the two angles in a right triangle brainly.com/question/4302397

Keywords: m∠a = 47°, m∠b, m∠c, and m∠d, 133°, vertical angles, supplementary, 180°, congruent

7 0

3 years ago

Read 2 more answers

Is 6 a solution to the equation? Show your work.

meriva

Answer:

yes I think 6 is a solution to the equation

7 0

2 years ago

Read 2 more answers

A house purchased for $216,500 gains 2% of its value every year. If this growth rate continues, in how many years will the house

NikAS [45]

Answer: 36 years

Step-by-step explanation:

You can use the Rule of 72 to calculate how long it might take the house to double in value.

The Rule of 72 works by dividing 72 by the interest rate as a whole number and the result will be a rough estimate of the time in years it will take for the investment to double in size:

= 72 / 2

= 36 years

3 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

ZC and ZD are vertical angles with mZC = -3x+58 and mZD= x - 2.

kogti [31]

Answer:

×=-2

0=×-2

×=-2

33333333

3 0

2 years ago