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2 years ago

6

In a Gallup Poll in 2003, 590 of 1,116 randomly selected adults aged 18 and older said they had too little time for relaxing or

doing nothing. Identify the upper level of a 95% confidence interval of the true proportion of adults who say they have too little time for relaxing.

1 answer:

White raven [17]2 years ago

8 0

Answer:

hope this helps

Step-by-step explanation:

what do you have to do

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PLEASE HELP THIS IS URGENT

Lisa [10]

X= 1, x = 5

-3|x - 3| = -6
———— —– divide both sides by -3
-3 -3 and remove the bars

x - 3 = 2
+3 +3
—————
x = 5⭐️

-3|1-3|➡️-3|-2|= -6

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2 years ago

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Help on number 6 i don't understand how to do it and i got corrections so i need help to do them.

Mademuasel [1]

The answer #6 is j. 100. He uses the treadmill for 40% so that means 60% of his exercise is not on the treadmill.

40%= 40 mins
60%= 60 mins
total: 100%= 100 mins.
Hope this helps!

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2 years ago

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What is NOT true about a direct proportion

SVEN [57.7K]

It’s graph must go through the origin ?

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2 years ago

Justin is 2 years older than one-third Norris's age Amy is four years younger than 2 times Justin's age to find a variable and w

djyliett [7]

So alyssa is 63 right? so 63/2/4/Ithird

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2 years ago

The owner of a computer repair shop has determined that their dailyrevenue has mean $7200 and standard deviation $1200. The dail

meriva

Answer:

The required probability is 0.0855

Step-by-step explanation:

Consider the provided information.

The daily revenue has mean $7200 and standard deviation $1200.

$\mu_{\bar x}=7200$

$\sigma=1200$

The daily revenue totals for the next 30 days will be monitored.

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x}=\frac{1200}{\sqrt{30}}=219.089$

As we know $Z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}$

Substitute $\bar x=7500, \mu_{\bar x}=7200\ and\ \sigma_{\bar x}=219.089$ in above formula.

$Z=\frac{7500-7200}{219.089}=1.3693$

From the standard normal table P( Z >1.3693) = 0.0855

Hence, the required probability is 0.0855

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2 years ago