All categories

2 years ago

What equation represents a population of 210 animals that decrease at an annual rate of 14%?

Mathematics

1 answer:

kumpel [21]2 years ago

8 0

So for this, we will be using exponential form, which is y = ab^x (a = initial value, b = growth/decay).

Since we start off with 210 animals, that is our a variable.

Next, since this is *decreasing* by 14%, you are to subtract 0.14 (14% in decimal form) from 1 to get 0.86. That will be your b variable.

Putting everything together, your equation is y = 210(0.86)^x

You might be interested in

A bag contains 100 marbles which are red, green, and blue. Suppose a student randomly selects a marble without looking, records

ra1l [238]

Answer:

35 red marbles

Step-by-step explanation:

So we can put the different color marbles in a ratio. So 7:2:11. If we add all of those up we get 20. We do 100/20 to get 5. Each number in the ratio is equivalent to 5 marbles. Then we do 7*5 to get 35. We can check our work by doing, (7*5) + (2*5) + (11*5) = 100. And it does equal 100 so the answer is correct.

6 0

2 years ago

Read 2 more answers

Michael is shipping his mother's birthday gift to her in a rectangular box. If the gift's dimensions are 2 inches long by 5 inch

stepan [7]

Answer:

112

Step-by-step explanation:

4 0

2 years ago

If m∠DEF = 117°, find the value of x

ollegr [7]

Answer:

x = 7

Step-by-step explanation:

Given:

∠DEF = 117°

∠DEG = (12x + 1)°

∠GEF = (5x - 3)°

Find:

value of x

Computation:

∠DEF = ∠DEG + ∠GEF

117° = (12x + 1)° + (5x - 3)°

117° = 17 x - 2

x = 7

4 0

2 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago

At a high school, the probability that a student is a senior is 0.25. The probability that a student plays a sport is 0.15. The

Serggg [28]

Use the definition of conditional probability:

$P(\text{sports}\mid\text{senior})=\dfrac{P(\text{sports AND senior})}{P(\text{senior})}$

We know that 0.10 of students belong to both categories, and that 0.25 of students are seniors, so

$P(\text{sports}\mid\text{senior})=\dfrac{0.10}{0.25}=0.40$

5 0

2 years ago