Which symbol makes the sentence true? − 9/16 ? -3/5 A.< B.> C.=

1 answer:

7 0

$- \frac{9}{16}= -\frac{45}{80} \\ \\ -\frac{3}{5} = -\frac{48}{80} \\ \\ -\frac{45}{80} >- \frac{48}{80} \\ \\ =>- \frac{9}{16} > -\frac{3}{5} 

 \\ \\ \framebox[1.1\width]{See u :)} \par$

You might be interested in

A candle burned at a steady rate. After 37 minutes, the candle was 11.2 inches tall. Eighteen minutes later,

irinina [24]

Answer:

i think the answer is 4.65 inches tall.

Step-by-step explanation:

i attempted it this may be wrong just wait for someone to confirm it.

3 0

3 years ago

Please help me I don’t understand what I’m supposed to do ! Unit 1 - Triangle Congruence

Luba_88 [7]

Answer:

i dont either

Step-by-step explanation:

6 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$