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3 years ago

What is 39 to 1 significant figure

Mathematics

2 answers:

wolverine [178]3 years ago

6 0

<h3>Answer: 40</h3>

Explanation

We can only keep one of the digits: either the tens digit or the units digit. If we kept only the units digit, then we go from 39 to 9 which is a very steep drop. If we kept the tens digit then we go from 39 to 30

The 0 in the "30" is not significant. It's just a placeholder to help us tell 30 apart from simply 3. The drop from 39 to 30 is much smaller (compared to from 39 to 9). But you can probably see that it's even better to bump up to 40 instead, due to the smaller change.

In short, 39 rounds up to 40 when rounding to one significant digit.

Veseljchak [2.6K]3 years ago

3 0

Which statement best completes the diagram?
Cultural features
of China
The Han are the
largest ethnic
group
Most people speak
Mandarin Chinese?
A. Its music is based mostly on Japanese traditions.
B. Its most popular art is critical of the government.
C. Different regions have their own unique cuisines.
D. Most people live in small farming villages.

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Find the area of the figure.

seraphim [82]

$\huge \bf༆ Answer ༄$

At first Divide the figure into two rectangles, I and Il

Area of figure l is ~

$\sf24 \times (40 - 30)$

$\sf24 \times 10$

$\sf240 \: \: ft {}^{2}$

Area of figure ll is ~

$\sf 18 \times 30$

$\sf540 \: ft{}^{2}$

Area of whole figure = Area ( l + ll )

that is equal to ~

$\sf240 + 540$

$\sf780 \: \: ft{}^{2}$

7 0

3 years ago

About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two

Flauer [41]

$p(A) = \frac{1}{2}$

$p(B) = \frac{1}{6}$

$p(C) = \frac{1}{6}$

$P(A | C)=\frac{1}{2}$

<u>Solution:</u>

The probability of an event is given as:

$\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}$

In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}

<em><u>First let us calculate p(A):</u></em>

The event is defined as: The sum on the two dice is even

Sum on two dice is even if and only if either both dice turn up odd or both even.

The odd outcomes in thowing a single die = 3 {1, 3, 5}

The even outcomes in throwing a single die = 3 {2, 4, 6}

The probability that both turn up odd is:

$\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

<em><u>Similarly, the probability that both turn up even is:</u></em>

$\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even

$\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Thus $p(A) = \frac{1}{2}$

<em><u>Let us calculate p(B):</u></em>

The event B is defined as: The sum on the two dice is at least 10

The total possible outcomes of two die is given as:

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :

atleast 10 means that sum can be 10 or greater than 10

{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}

Here favourable outcomes = 6

Total number of outcomes = 36

Hence, the probability that the sum of the two dice will be at least 10 is:

$\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}$

Thus $p(B) = \frac{1}{6}$

<em><u>Let us calculate p(C):</u></em>

The event C is defined as: The red die comes up 5

Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

$\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}$

Thus $p(C) = \frac{1}{6}$

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

$\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}$

$p(A \cap C)=\frac{3}{36}=\frac{1}{12}$

$P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$

Thus $P(A | C)=\frac{1}{2}$

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4 years ago

After filling a gas tank the odometer read 82,803.6 After the next filling it read 83,034.6 it took 16.5 gallons to fill the tan

ELEN [110]

83,034.6 - 82,803.6 = 231 miles

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3 years ago

I want answer for this quiz question

damaskus [11]

Answer:

ummm since when did they add letters into math we ain't learning the alphabet it's math! like what

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2 years ago

Mrs. Morales wrote a test with 16 questions covering spelling and vocabulary. Spelling questions (x) are

fiasKO [112]

Answer:

Step-by-step explanation:

X+Y = 15

<h3>Sum of two types of question is 15</h3>

x+y-x = -x+15

5(x)+10(y)=100

5x+10y=100

<h3>Total no. if points is 100</h3>

5x+10y-5x=-5x+100

10y= -5x+100

y=5/10x + 100/10

y= -1/2x+10

<h3>Slope intercept form</h3>

(10,5)

8 0

3 years ago

What is 39 to 1 significant figure​

What is 39 to 1 significant figure