The answer=ajshabeel

A certain culture of year increases by 50% every three hours. A scientist places 9 grams of yeast in a culture dish. Write the e

miskamm [114]

Answer: a) Explicit formula : $f(x)=9(1.5)^{3x}$

b) Recursive formula : $f(x)=9(1.5)^{3x}$ and $f(x+1)=f(x)\times 1.5^3$

Step-by-step explanation:

Since we have given that

Number of grams of yeast in a culture dish initially = 9 grams

According to question, a certain culture of year increases by 50 % every three hours.

So, the explicit formula will be

$f(x)=9(1+\frac{50}{100})^{3x}\\\\f(x)=9(1+0.5)^{3x}\\\\f(x)=9(1.5)^{3x}$

And the recursive formula will be

$f(x)=9(1.5)^{3x}\\\\f(x+1)=9(1.5)^{3(x+1)}\\\\f(x+1)=9(1.5)^{3x+3}\\\\f(x+1)=9(1.5)^{3x}.(1.5)^3\\\\f(x+1)=f(x)\times 1.5^3$

Hence, a) Explicit formula : $f(x)=9(1.5)^{3x}$

b) Recursive formula : $f(x)=9(1.5)^{3x}$ and $f(x+1)=f(x)\times 1.5^3$

6 0

3 years ago

Help with 2 3 4 and 5 please!

pantera1 [17]

Answer:

Number 2 is A

Number 3 is (5, -4)

Number 4 is (5, -4)

Number 5 is (4, 2)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Suppose 46% of politicians are lawyers. If a random sample of size 662 is selected, what is the probability that the proportion

Svet_ta [14]

Answer:

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$