Which equation could be used to find the length of the hypotenuse?
А
С
5 cm
С
B
8 cm

Answer: 5/55

Step-by-step explanation:

1/11 x 5 = 5/55

So, the fifth equivalent fraction to 1/11 is 5/55.

The 5th equivalent fraction should be [tex]5\div 55[/tex]

Since the fraction is [tex]1\div 11[/tex]

So here the 5th equivalent should be

[tex]= 1\div 11 \times 5\div 5[/tex]

= [tex]5\div 55[/tex]

Here 5 represent the numerator and 55 represent the denominator.

Therefore, we can concluded that The 5th equivalent fraction should be [tex]5\div 55[/tex]

Learn more about fraction here: https://brainly.com/question/1786648

Answer:

the 95th percentile for the sum of the rounding errors is 21.236

Step-by-step explanation:

Let consider X to be the rounding errors

Then; [tex]X \sim U (a,b)[/tex]

where;

a = -0.5 and b = 0.5

Also;

Since The error on each loss is independently and uniformly distributed

Then;

[tex]\sum X _1 \sim N ( n \mu , n \sigma^2)[/tex]

where;

n = 2000

Mean [tex]\mu = \dfrac{a+b}{2}[/tex]

[tex]\mu = \dfrac{-0.5+0.5}{2}[/tex]

[tex]\mu =0[/tex]

[tex]\sigma^2 = \dfrac{(b-a)^2}{12}[/tex]

[tex]\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}[/tex]

[tex]\sigma^2 = \dfrac{(0.5+0.5)^2}{12}[/tex]

[tex]\sigma^2 = \dfrac{(1.0)^2}{12}[/tex]

[tex]\sigma^2 = \dfrac{1}{12}[/tex]

Recall:

[tex]\sum X _1 \sim N ( n \mu , n \sigma^2)[/tex]

[tex]n\mu = 2000 \times 0 = 0[/tex]

[tex]n \sigma^2 = 2000 \times \dfrac{1}{12} = \dfrac{2000}{12}[/tex]

For 95th percentile or below

[tex]P(\overline X < 95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95[/tex]

[tex]P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95[/tex]

[tex]P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95[/tex]

[tex]\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95[/tex]

[tex]\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05[/tex]

From Normal table; Z >   1.645 = 0.05

[tex]\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645[/tex]

[tex]{P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}[/tex]

[tex]{P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }[/tex]

[tex]\mathbf{P_{95} = 21.236}[/tex]

the 95th percentile for the sum of the rounding errors is 21.236

hello,

the first term is 250 so this is the initial invested amount

[tex](1+\dfrac{0.08}{12})^{12}=(1+\dfrac{8\%}{12})^{12}[/tex]

is to compute 8% annual interest compounded monthly (there are 12 months in a year)

and then multiply by 4 means that it is computed for 4 years so

finally the answer is

$250 is invested at 8% annual interest compounded monthly for 4 years

hope this helps

Step-by-step explanation:

1/3 and 5/2 can be shown as:

points with 1/6 interval on the number line:

0, 1/6, 2/6, 3/6, 4/6, ..., 15/6

Answer:

0.03 feet

Step-by-step explanation:

d = 0.03r² + r

When d = 0: 0.03r² + r = 0

r(0.03r + 1) = 0

∴ r = 0

When r = 0: d = 0.03 feet