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igor_vitrenko [27]
3 years ago
9

Help me please (look the picture)

Mathematics
1 answer:
umka2103 [35]3 years ago
4 0
You add 1 to the numerator each time
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Round 314550 to 1 significant figure​
Sauron [17]
300000 all you had to do was round the number down after the 1 significant figure

you round down with
1,2,3,4

and round up with
5,6,7,8,9

so if it asked for 78453
to 2 significant figures you count the numbers and round after that
so 78|453 and since 4 is the next number and it is less that 5 we round down which means it becomes 78000 with is lower than 78453

but if we had 0.0247
and it asked for this number to 2 significant figures all you need to do is count the SIGNIFICANT figures significant means anything with value so that is 1,2,3,4,5,6,7,8,9,10.... 0 has no value therefore we ignore 0 and start counting at the first figure larger than 0
if we go back to 0.0247 that would be the 2 so 2 significant figure of 0.0247 would equal
0.024|7
7 is larger than 5 so we round up to
0.025
tadaa i hope that helped
8 0
3 years ago
Read 2 more answers
Plz, Help me with this question.
Alexxx [7]

Answer:

B

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Find the equation of a line that contains the points (6, -7) and (8,5). Write the equation in slope-intercept form.
Anna [14]

Answer:

y = 6x - 43

Step-by-step explanation:

(6, -7) and (8,5)

m=(y2-y1)/(x2-x1)

m=(5 + 7)/(8 - 6)

m= 12/2

m = 6

y - y1 = m(x - x1)

y + 7 = 6(x - 6)

y + 7 = 6x - 36

y = 6x - 43

6 0
3 years ago
The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same
masha68 [24]

Answer: The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.

Step-by-step explanation:

Let x and y area the random variable that represents the heights of women and men.

Given : The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches.

i.e. \mu_1 = 64   \sigma_1=2.7

Since , z=\dfrac{x-\mu}{\sigma}

Then, z-score corresponds to  a woman 6 feet tall (i.e. x=72 inches).

[∵  1 foot = 12 inches , 6 feet = 6(12)=72 inches]

z=\dfrac{72-64}{2.7}=2.96296296\approx2.96

Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.

i.e. \mu_2 = 69.3   \sigma_2=2.8

Then, z-score corresponds to a man 5'10" tall (i.e. y =70 inches).

[∵  1 foot = 12 inches , 5 feet 10 inches= 5(12)+10=70 inches]

z=\dfrac{70-69.3}{2.8}=0.25

∴ The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.

6 0
3 years ago
Pls help will give Brainliest if right, Don’t guess
soldier1979 [14.2K]
It’s a every minute he scores one more point
7 0
3 years ago
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