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3 years ago

14

The fish population in a local stream is decreasing at a rate of 5% per year. The original population was 75,000 fish. What is t

he population of the fish pond after 7 years?

1 answer:

ZanzabumX [31]3 years ago

4 0

Answer:

52,375

Step-by-step explanation:

that is what i got

You might be interested in

F(7) if f(x)=-3x-2

Paul [167]

Answer:

f(7) = -23

Step-by-step explanation:

f(x) = -3x - 2

f(7) = -3(7) - 2

f(7) = -21 - 2 = -23

4 0

2 years ago

You are selling candy bars for the school to

Strike441 [17]

Answer:

200 bars

Step-by-step explanation:

$300 - $50 = the amount of money you still need, $250

250 dollars/$1.25 per bar = 200 bars you need to sell to get 300 dollars

7 0

3 years ago

Chad gets an annual salary of 25,000. He and his family spend 3500 per year on food. What percent of his salary is spent on food

Advocard [28]

Answer:

$\frac{x}{100}$ x 25000 = 3500

$x$ x 250 = 3500

x = 3500/250

x = 350/25

x = 14%

7 0

3 years ago

What is 91/22 in decimal?

Lemur [1.5K]

<span>4.13 (continues) is the answer</span>

5 0

3 years ago

A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0

3 years ago