30% were sponge cakes because you do 1/5=0.2 or20%+50%=70% and 100%-70%=30% so 30% is your answer.
Answer:
To determine the number of real number solutions of as system of equations in which one equation is linear and the other is quadratic
1) Given that there are two variables, x and y as an example, we make y the subject of the equation of the linear equation and substitute the the expression for y in x into the quadratic equation
We simplify and check the number of real roots with the quadratic formula,
for quadratic equations the form 0 = a·x² - b·x + c
Where b² > 4·a·c there are two possible solutions and when b² = 4·a·c equation there is only one solution.
Step-by-step explanation:
Percent increase is:
Change/original * 100
(29975-24400)/24400
5575/24400= .228 * 100=
22.8%
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:
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
Answer:
1.4in
Step-by-step explanation:
Length of Photo = 4in
Width of Photo = 3in
Unknown:
Value of X = ?
Solution:
Follow these steps:
Area of a rectangle = l x w
Since the photo is a rectangle; area of photo:
Area of photo = 4in x 3in = 12in²
For the area of the ad;
Length of ad = 4 + x
Width of ad = 3 + x
Given that,
the area of the photo =
area of ad
12in² =
area of ad
Area of ad = 24in²
Area of the ad;
(4 + x) (3 + x) = 24
12 + 4x + 3x + x² = 24
12 + 7x + x² = 24
x² + 7x = 24 - 12
x² + 7x = 12
x² + 7x - 12 = 0
Using the almighty formula where
a = 1, b = 7 and c = -12
x = 
x =
or ![\frac{-7 - \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}](https://tex.z-dn.net/?f=%5Cfrac%7B-7%20-%20%5Csqrt%5B%5D%7B-7%5E%7B2%7D%20-%204x1x-12%20%7D%20%7D%7B2x1%7D)
x = 1.4 or -8.4
therefore the answer is 1.4in
x is 1.4in