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3 years ago

6

the population of center city is modeled by exponential function f, where x is the number of years after the year of 2015. the g

raph of f is shown on the grid. which inequality best represents the range of f in this situation?

1 answer:

lora16 [44]3 years ago

5 0

Answer:

the Answer is B) y >= to 250,000

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What’s the slope?!??????

Nataly_w [17]

Answer:

y=mx+b

Step-by-step explanation:

4 0

3 years ago

A sample of 2,000 people yielded P(hat) = .52. What is the variance of the population proportion?

lys-0071 [83]

Answer:

d. .00012 or 0.00012

Step-by-step explanation:

The variance of the population proportion formula is given as:

p(1 - p)/n

Where n = sample size

p = population proportion or p(hat)

n = 2000 people

p = 0.52

Variance of the population proportion = 0.52(1 - 0.52)/ 2000

= 0.52 × 0.48 /2000

= 0.2496 /2000

= 0.0001248

Therefore the Variance of the population proportion is approximately = 0.00012

3 0

3 years ago

Given that S^3 on the bottom 1 (e^x)dx=(e^3)-e use the properties of integrals and this result to evaluate S^3 on the bottom 1 (

scZoUnD [109]

Remember that
1) For functions $f(x)$ and $g(x)$ , $\int {f(x) + g(x)} \, dx = \int {f(x)} \, dx + \int {g(x)} \, dx$
2) For a function $f(x)$ and a constant $c$ , $\int {cf(x)} \, dx = c \int {f(x)} \, dx$

Using these two properties of integrals, and the fact that $\int\limits^3_1 {e^x} \, dx = e^3 - e$ , we can see that

$\int\limits^3_1 {5e^x - 1} \, dx$
$= \int\limits^3_1 {5e^x} \, dx - \int\limits^3_1 1} \, dx$
$= 5 \int\limits^3_1 {e^x} \, dx - \int\limits^3_1 1} \, dx$
$= 5(e^3 - e) - \left.x\right|_1^3$
$= 5e^3 - 5e - (3 - 1)$
$= \bf 5e^3 - 5e - 2$

6 0

3 years ago

A bag contains 26 tiles with letters A through Z. What is the probability of drawing 2 vowels without replacement

pav-90 [236]

1st draw: 5/26

2nd draw 4/25

probability of 2 vowels = 5/26* 4/25

rearrange to make the math easier

= 5/25* 4/26

= 1/5 * 2/13

=2/65

5 0

4 years ago

In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped

umka21 [38]

Answer:

The 95% confidece estimate for how much a typical parent would spend on their child's birthday gift is between $37.47 and $46.53.

Step-by-step explanation:

The results were roughly normal, so we can find the normal confidence interval.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{12}{\sqrt{27}} = 4.53$

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 4.53 = $37.47.

The upper end of the interval is the sample mean added to M. So it is 42 + 4.53 = $46.53.

The 95% confidece estimate for how much a typical parent would spend on their child's birthday gift is between $37.47 and $46.53.

5 0

4 years ago