Ask question

Ask question

All categories

3 years ago

9

If the population of a town was experiencing uninhibited decay of 2% per year, how long would it take the population to decrease

to half? Round to the nearest tenth.
48.6 years
26.4 years
15.1 years
34.7 years

1 answer:

hichkok12 [17]3 years ago

7 0

Answer:

replied in the wrong place whoops the answer is 34.7

Step-by-step explanation:

You might be interested in

Adriano loves anime and collects toys from his favorite series. each month he gains 4 toys. if adriano has 3 toys when he starts

skad [1K]

The equation for this would be 4x+3=y where x is months and y is your amount of toys after x months.
plug in 4 for x -> 4(4)+3=y
solve -> 16+3=y
y=19

8 0

3 years ago

Read 2 more answers

What is the image point of (-6,3) after a translation right 5 units and down 3 units?

kobusy [5.1K]

Answer:
(-1,0)
step-by-step explanation:
(-6,3)
right 5
(-1,3)
down 3
(-1,0)

6 0

3 years ago

HELP PLZ I GIVE YOU BRAINLIEST IF YOU GET CORRECT PLZ I BEG YOU

DiKsa [7]

Answer:

Original number = 38

Step-by-step explanation:

10*x + y

x = y - 5

10y + x = 2 (10x + y) + 7

10y + y - 5 = 2(10(y - 5) + y) + 7

11y - 5 = 2(10y - 50 + y ) + 7

11y - 5 = 2(11y - 50) + 7

11y - 5 = 22y - 100 + 7

11y - 5 = 22y - 93

11y + 88 = 22y

88 = 11y

y = 8

x = y - 5

x = 8 - 5

x = 3

So the original number is 38. Does that work?

3 is 5 less than 8.

83 = 2*38 + 7

83 = 76 + 7

83 = 83. Yes it works.

8 0

3 years ago

Which of the following represents the most accurate estimation of 664 - 127?

Arlecino [84]

The correct answer is 537

3 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago