To solve this we are going to use the formula fro the force applied to a spring:
where
is the spring constant
is the extension
Since we know the
, we can replace that in our formula and solve for
:
where
is the acceleration
is the spring constant
is the extension
is the mass
We know for our problem that
,
, and
. So lets replace those values in our formula to find
:
We can conclude that the acceleration of the block when s=0.4m is
.
Answer:
The sample consisting of 64 data values would give a greater precision.
Step-by-step explanation:
The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:
So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).
That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.
Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.
The two sample sizes are:
<em>n</em>₁ = 25
<em>n</em>₂ = 64
The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.
Width for n = 25:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B25%7D%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Width for n = 64:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B64%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Thus, the sample consisting of 64 data values would give a greater precision
First, lets create a equation for our situation. Let
be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span>
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:
. So lets replace the values in our formula to find the average of our function:
![\frac{25(10)+100-[25(2)+100]}{10-2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%2810%29%2B100-%5B25%282%29%2B100%5D%7D%7B10-2%7D%20)
We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
Answer:
Step-by-step explanation:
Given:
°
From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.
Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,
From the Δ BCD, which is a right angled triangle,
Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.
Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles 
and EF is a tangent to the circle at point C.
Therefore, 
Again, using the same theorem as above,
Hence, all the angles are as follows:
