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Musya8 [376]
3 years ago
7

You measure 20 dogs' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 11.5 ou

nces. Based on this, construct a 95% confidence interval for the true population mean dog weight.
Mathematics
1 answer:
Assoli18 [71]3 years ago
4 0

95% confidence interval would be (58.96, 69.04).

Step-by-step explanation:

Since we have given that

Number of dogs' weight = 20

Mean = 64 ounces

Standard deviation = 11.5 ounces

We need to find the 95% confidence interval.

So, z = 1.96

so, interval would be

\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=64\pm 1.96\times \dfrac{11.5}{\sqrt{20}}\\\\=64\pm 5.04\\\\=(64-5.04,64+5.04)\\\\=(58.96,69.04)

Hence, 95% confidence interval would be (58.96, 69.04).

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if Dakota earned $15.75 in interest in account A and $28.00 in interest in account B after 21 months. if the simple interest rat
NISA [10]

keeping in mind that 21 months is more than a year, since there are 12 months in a year,  then 21 months is really 21/12 years.


\bf ~~~~~~ \stackrel{\textit{account A}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &15.75\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 15.75=P(0.03)\left( \frac{7}{4} \right)\implies \cfrac{15.75}{(0.03)\left( \frac{7}{4} \right)}=P\implies \boxed{300=P}


\bf ~~~~~~ \stackrel{\textit{account B}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &28\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 28=P(0.049)\left( \frac{7}{4} \right)\implies \cfrac{28}{(0.049)\left( \frac{7}{4} \right)}=P\implies \boxed{326.53\approx P}


so, clearly, you can see who's greater.

3 0
3 years ago
the map shows Henry's town . Each unit represents 1 kilometer. After school, Henry walks to the library. How far does he walk?​
ivann1987 [24]

Answer:

Henry walks 7 kilometres from school to library

Step-by-step explanation:

Given:

1 unit = 1 Kilometre

To Find:

The distance between the library and the school

Solution:

The map below shows that  Henry has walked from

-2 coordinate in the x axis to coordinate 5  along the same x axis

So he has walked a total of 7 units

Since we know that 1 unit  = 1 Kilometre,

7 units will be

=> 7 x 1

=> 7 Kilometres

3 0
3 years ago
Read 2 more answers
Area of the base=
umka21 [38]

Answer:

16 pi

Step-by-step explanation:

The base is the circle

A = pi r^2

A = pi (4)^2

A = 16 pi

3 0
2 years ago
Read 2 more answers
The answers to (a)-(d) work is not needed just answers
Debora [2.8K]
A)  n(a) = 25 + 15 = 40
b) n(b) = 42+15 =57
c) n (a U b) = 25 +15+ 42 = 82 
d) n (a') = 42
e) n(b') = 25
f)  n(a intersection b)' =  82-15 = 67 
g) n (a U b)' =     14 
h)  n (a'  intersection  b') =  14
i)  n (a' U b') = 42 + 25 +14 = 81 


hope i helped a bit .. :-)
6 0
3 years ago
Solve the triangle.
Angelina_Jolie [31]
Using the sine law of triangles to solve for angle C:

b / sin B = c / sin C

C = arcsin (c * sin B / b)
C = 30.7<span>°

Since interior angles of a triangle always add up to 180</span><span>°, we can use this to solve for angle A:

angle A = 180</span>° - 73° - 30.7<span>°
angle A = 76.3</span><span>°

Having solved for angle A, we can solve for side a using the sine law.

a / sin A = b / sin B

a = b * sin A / sin B
a = 15.2

Therefore, the correct answer is B.</span>
3 0
2 years ago
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