Answer:
False
Step-by-step explanation:
If we were to flip the solid on the left so its parallel to the solid on the right, we would be able to compare the two more easier.
We can see that the right solid has dimensions of:
L = 1 cm
W = 3 cm
H = 5 cm
The left solid has dimensions of:
L = 1
W = 2
H = 7
If we were to add these all up, they would not equal.
R: 1 + 3 + 5 = 9
L: 1 + 2 + 7 = 10
Full year membership=$325
Monthly membership= (30)(12)=360+ 100 upfront fee.
Full year membership =$325
Mounted membership =$460
Full year membership is $100 cheaper.
Yes, we claim that the two ratios are proportional.
<h3>Proportion</h3>
Proportion is an equality between two ratios. These ratios are made up of two real integers. In a proportion, when there is an equality, both ratios have to be equivalent to the other. Equivalent ratio is when both division results are equal.
— To find out if the equality between these two given ratios is true, let's just: divide the numerator by the denominator.
3/8 = 3 ÷ 8 = <u>0,375</u>
24/64 = 24 ÷ 64 = <u>0,375</u>
Therefore, we see that the results of these ratios are the same. So these two ratios are proportional. 3/8 = 24/64
Answer:
We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The precision of the confidence interval depends on the margin of error ME = Zcritical * Sqrt[(p(1-p)/n]
In this Zcritical value is in the numerator. Z critical decreases as Confidence level decreases. (Zc for 99% = 2.576, Zc for 95% is 1.96, Zc for 90% = 1.645). Therefore decreasing the Confidence level decreases ME.
Also we see that sample size n is in the denominator. So the ME decreases as sample size increases.
Therefore, We can select a lower confidence level and increase the sample size.
Answer:
The ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.
Step-by-step explanation:
Let the Area of smaller watch face be ![A_1](https://tex.z-dn.net/?f=A_1)
Also Let the Area of Larger watch face be ![A_2](https://tex.z-dn.net/?f=A_2)
Also Let the radius of smaller watch face be ![r_1](https://tex.z-dn.net/?f=r_1)
Also Let the radius of Larger watch face be ![r_2](https://tex.z-dn.net/?f=r_2)
Now given:
![\frac{A_1}{A_2} =\frac{16}{25}](https://tex.z-dn.net/?f=%5Cfrac%7BA_1%7D%7BA_2%7D%20%3D%5Cfrac%7B16%7D%7B25%7D)
We need to find the ratio of the radius of the smaller watch face to the radius of the larger watch face.
Solution:
Since the watch face is in circular form.
Then we can say that;
Area of the circle is equal 'π' times square of the radius 'r'.
framing in equation form we get;
![A_1 = \pi {r_1}^2](https://tex.z-dn.net/?f=A_1%20%3D%20%5Cpi%20%7Br_1%7D%5E2)
![A_2 = \pi {r_2}^2](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cpi%20%7Br_2%7D%5E2)
So we get;
![\frac{A_1}{A_2}= \frac{\pi {r_1}^2}{\pi {r_2}^2}](https://tex.z-dn.net/?f=%5Cfrac%7BA_1%7D%7BA_2%7D%3D%20%5Cfrac%7B%5Cpi%20%7Br_1%7D%5E2%7D%7B%5Cpi%20%7Br_2%7D%5E2%7D)
Substituting the value we get;
![\frac{16}{25}= \frac{\pi {r_1}^2}{\pi {r_2}^2}](https://tex.z-dn.net/?f=%5Cfrac%7B16%7D%7B25%7D%3D%20%5Cfrac%7B%5Cpi%20%7Br_1%7D%5E2%7D%7B%5Cpi%20%7Br_2%7D%5E2%7D)
Now 'π' from numerator and denominator gets cancelled.
![\frac{16}{25}= \frac{{r_1}^2}{{r_2}^2}](https://tex.z-dn.net/?f=%5Cfrac%7B16%7D%7B25%7D%3D%20%5Cfrac%7B%7Br_1%7D%5E2%7D%7B%7Br_2%7D%5E2%7D)
Now Taking square roots on both side we get;
![\sqrt{\frac{16}{25}}= \sqrt{\frac{{r_1}^2}{{r_2}^2}}\\\\\frac{4}{5}= \frac{r_1}{r_2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B16%7D%7B25%7D%7D%3D%20%5Csqrt%7B%5Cfrac%7B%7Br_1%7D%5E2%7D%7B%7Br_2%7D%5E2%7D%7D%5C%5C%5C%5C%5Cfrac%7B4%7D%7B5%7D%3D%20%5Cfrac%7Br_1%7D%7Br_2%7D)
Hence the ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.