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wariber [46]
4 years ago
5

In a random sample of 13 people. The mean length of stay at a hospital was 6.2 days. Assume the population standard deviation is

1.7 days and the length of stay are normally distributed. Construct a 99% and a 95% confidence interval.
Mathematics
2 answers:
Alborosie4 years ago
7 0

Answer:

Do you know this answer?

Step-by-step explanation:

Theseus forgives Arcite and Palamon and then proposes a way to let Destiny determine who will win Emily's hand. What doe he suggest? What rules must be followed?

matrenka [14]4 years ago
4 0

Answer:

We have 99% and 95% confidence interval to be

7.1241 Or 5.276

Step-by-step explanation:

Given that n = 13

Mean = 6.2

Standard deviation = 1.7

Therefore to construct confidence interval for 99% and 95%.

Z score for 95% confidence level is 1.96

We have that:

6.2 +/- 1.96(1.7/√13)

6.2 +/- 0.9241

6.2+0.9241 or 6.2-0.9241

7.1241 Or 5.276

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If tan θ=3\4 , find csc
Rom4ik [11]

In a rectangular triangle \tan\theta\dfrac{opposite}{adjacent}.

We have \tan\theta=\dfrac{3}{4}

Therefore opposite = 3uints and adjacent = 4units.

\csc\theta=\dfrac{hypotenuse}{opposite}

We need the length of hypotenuse.

Use the Pythagorean theorem:

h^2=3^2+4^2\\\\h^2=9+16\\\\h^2=25\to h=\sqrt{25}\to h=5units

Substitute:

\csc\theta=\dfrac{5}{3}

3 0
4 years ago
Find the H.C.F of 144,180,240,300​
aliina [53]
Answer:
GCF = 12

for the values 144, 180, 240, 300

Solution by Factorization:

The factors of 144 are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144

The factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180

The factors of 240 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240

The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300

Then the greatest common factor is 12.
8 0
3 years ago
Read 2 more answers
What is the value of 5 units as a mixed number
masha68 [24]

5integer 0/1 is the mixed fraction of 5 units

8 0
3 years ago
The average weight of a particular box of crackers is 28.0 ounces with a standard deviation of 0.9 ounce. The weights of the box
Juliette [100K]

Answer:

a. 97.72%

Step-by-step explanation:

The weights of boxes follows normal distribution with mean=28 ounce and standard deviation=0.9 ounces.

a. We have to calculated the percentage of the boxes that weighs more than 26.2 ounces.

Let X be the weight of boxes. We have to find P(X>26.2).

The given mean and Standard deviations are μ=28 and σ=0.9.

P(X>26.2)= P((X-μ/σ )> (26.2-28)/0.9)

P(X>26.2)= P(z> (-1.8/0.9))

P(X>26.2)= P(z>-2)

P(X>26.2)= P(0<z<∞)+P(-2<z<0)

P(-2<z<0) is computed by looking 2.00 in table of areas under the unit normal curve.

P(X>26.2)=0.5+0.4772

P(X>26.2)= 0.9772

Thus, the percent of the boxes weigh more than 26.2 ounces is 97.72%

5 0
3 years ago
DB _____ RT Choose the relationship symbol to make a true statement. <br> &gt; <br> =
Nikitich [7]
DB = RT
We know this to be true because both have one “tally” mark indicating that they are congruent (same)
8 0
3 years ago
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