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3 years ago

6.

Mathematics

1 answer:

gladu [14]3 years ago

5 0

Answer:

Hope this is correct

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A newspaper found that, on average, 7.6% of people stay overnight in the hospital with a 1.5% margin of error. Construct a 95% c

tensa zangetsu [6.8K]

Given:
p = 7.6% = 0.076, the percentage of people who stay overnight at the hospital.
E = 1.5% = 0.015, margin of error
95% confidence interval.

The standard error is
Es = $\sqrt{ \frac{p(1-p)}{n} }$
where
n = the sample size.

The margin of error is
$E=z^{*}E_{s}$
where
z* = 1.96 at the 95% confidence level.

Because the margin of error is given, there is no need to calculate it.
The 95% confidence interval is
p +/- E = 0.076 +/- 0.015 = (0.061, 0.091) = (6.1%, 9.1%)

Answer:
The 95% confidence interval is between 6.1% and 9.1%.

3 0

3 years ago

What is 6:12 in simplest form? <br><br> A. 2:1 B. 12:6 C. 2 D. 1:2

AURORKA [14]

<span>the correct answer is D. 1:2</span>

4 0

3 years ago

Read 2 more answers

What 2 numbers whose sum is 105.7 the numbers are 19, 85.2, 533, 571, 88.2, 525, 20, 17.5, 400, 261, 20.5 ,125, 7, 23, 901, 30

givi [52]

To make it quickly we can consider only those numbers which will give us a decimal of xxx.7. There are only a few numbers like this: 85.2, 88.2, 20.5.

I tried adding 85.2 to 17.5 and it gave me 102.5 - it was close but not close enough :)

So I picked 85.2 + 20.5 = 105.7 and voila :)

6 0

3 years ago

A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the

nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

$\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1$

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

$\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1$

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

$\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883$

So, the horizontal distance from the center is 49.3883 feet

8 0

3 years ago

Sandra has taken 3 Test this Quarter. She has scored an 78%, 76%, and a 93%. Sandra wants to earn at least and 85% for the Quart

algol [13]

Answer:

I think at least a 86% hope this helps, hope i am right

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers