Answer:
10 cm²
Step-by-step explanation:
The area of a square = s², where s is the measure of the side
To calculate the side use Pythagoras' identity
Consider the lower side with right triangle legs 3 and 1, then
side² = 3² + 1² = 9 + 1 = 10 ( take the square root of both sides )
side = , thus
area = ( )² = 10 cm²
Answer:
73.5%
Step-by-step explanation:
The question is incomplete, the complete question is shown in the image attached.
Half life (t1/2) = 9 days
Time taken (t) = 4 days
From;
0.693/t1/2 = 2.303/t log (No/N)
Substituting values;
0.693/9 = 2.303/4 log(No/N)
0.693/9 × 4/2.303= log(No/N)
2.772/20.727= log(No/N)
Antilog (2.772/20.727)=No/N
1.36 = No/N
Therefore;
N/No = (1.36)^-1
N/No = 0.735
Percentage remaining after 4 days= 0.735× 100= 73.5%
Answer:
Step-by-step explanation:
Hello!
To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.
To test this claim two samples were taken:
Sample 1
X: battery lifespan of a unit of the new product (min)
n= 93 units of the new product
mean battery life X[bar]= 8:53hs= 533min
S= 84 min
Sample 2
X: battery lifespan of a unit of the leading product (min)
n= 102 units of the leading product
mean battery life X[bar]= 5:40 hs = 340min
S= 93 min
The population variances of both variances are unknown and distinct.
To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:
H₀: μ₁ - μ₂ ≤ 120
H₁: μ₁ - μ₂ > 120
Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:
Z≈N(0;1)
I hope this helps!