2 years ago

Help!!! (Tell me if you can’t see it clearly)

1 answer:

meriva2 years ago

3 0

I believe it’s option c!

You might be interested in

oksano4ka [1.4K]

Answer:

I think 8/10 is greater?

3 0

3 years ago

GalinKa [24]

6/9 is in 2/3 all you do is multiply numerator and denominator by 3

3 0

3 years ago

lukranit [14]

Sofia would have $34.75 left

4 0

3 years ago

lukranit [14]

Answer:

What kind of problem?

Step-by-step explanation:

7 0

3 years ago

WINSTONCH [101]

Answer: The number of first-year residents she must survey to be 95% confident= 263

Step-by-step explanation:

When population standard deviation ( $\sigma$ ) is known and margin of error(E) is given, then the minimum sample size (n) is given by :-

$n=(\dfrac{z^*\sigma}{E})^2$ , z* = Two-tailed critical value for the given confidence interval.

For 95% confidence level , z* = 1.96

As, $\sigma$ = 8.265, E = 1

So, $n= (\dfrac{1.96\times8.265}{1})^2 =(16.1994)^2\\\\= 262.42056036\approx263\ \ \ [\text{Rounded to the next integer}]$

Hence, the number of first-year residents she must survey to be 95% confident= 263

7 0

3 years ago