Answer:
>
Step-by-step explanation:
1 - 3 ?? -4
-2 ?? -4
The appropriate symbol for this ?? relation is >.
1 -3 > -4
Answer:
24%
Step-by-step explanation:
60/250=24
Answer:
Step-by-step explanation:
Part 1:
Let
Q₁ = Amount of the drug in the body after the first dose.
Q₂ = 250 mg
As we know that after 12 hours about 4% of the drug is still present in the body.
For Q₂,
we get:
Q₂ = 4% of Q₁ + 250
= (0.04 × 250) + 250
= 10 + 250
= 260 mg
Therefore, after the second dose, 260 mg of the drug is present in the body.
Now, for Q₃ :
We get;
Q₃ = 4% of Q2 + 250
= 0.04 × 260 + 250
= 10.4 + 250
= 260.4
For Q₄,
We get;
Q₄ = 4% of Q₃ + 250
= 0.04 × 260.4 + 250
= 10.416 + 250
= 260.416
Part 2:
To find out how large that amount is, we have to find Q₄₀.
Using the similar pattern
for Q₄₀,
We get;
Q₄₀ = 250 + 250 × (0.04)¹ + 250 × (0.04)² + 250 × (0.04)³⁹
Taking 250 as common;
Q₄₀ = 250 (1 + 0.04 + 0.042 + ⋯ + 0.0439)
= 2501 − 0.04401 − 0.04
Q₄₀ = 260.4167
Hence, The greatest amount of antibiotics in Susan’s body is 260.4167 mg.
Part 3:
From the previous 2 components of the matter, we all know that the best quantity of the antibiotic in Susan's body is regarding 260.4167 mg and it'll occur right once she has taken the last dose. However, we have a tendency to see that already once the fourth dose she had 260.416 mg of the drug in her system, that is simply insignificantly smaller. thus we will say that beginning on the second day of treatment, double every day there'll be regarding 260.416 mg of the antibiotic in her body. Over the course of the subsequent twelve hours {the quantity|the quantity|the number} of the drug can decrease to 4% of the most amount, that is 10.4166 mg. Then the cycle can repeat.
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!