What is the value of x?

Digiron [165]

Answer: 1211.6585 years

Step-by-step explanation:

The equation for exponential growth is: $P=P_oe^{kt}$

P: final population
P₀: initial population
k: rate of decay (or growth)
t: time

Use the half life information to find k:

$\dfrac{1}{2}P_o=P_oe^{k(800)}\\\\\\\dfrac{1}{2}=e^{800k}\qquad \rightarrow \qquad \text{divided both sides by}\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=800k\qquad \rightarrow \qquad \text{applied ln to both sides}\\\\\\\dfrac{ln(\frac{1}{2})}{800}=k\qquad \rightarrow \qquad \text{divided both sides by 800}\\\\\\\large\boxed{-0.000867=k}$

Next, input the information (65% decayed = 35% remaining) and the k-value to find your answer.

$.35P_o=P_oe^{-0.000867t}\\\\\\.35=e^{-0.000867t}\\\\\\ln(.35)=-0.000867t\\\\\\\dfrac{ln(.35)}{-0.000867}=t\\\\\\\large\boxed{1211.6585=t}$

5 0

3 years ago

Please help with these questions. Will mark brainliest!

andrew11 [14]

Answer:

1, pre image 2, transformation

Step-by-step explanation:

6 0

3 years ago

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

boyakko [2]

Answer:

$y^2 + 8y + 16$

Step-by-step explanation:

$(6y^2 + 2y + 5) - (5y^2 - 6y - 11) =$

The first set of parentheses is unnecessary, so it can just be removed.

$= 6y^2 + 2y + 5 - (5y^2 - 6y - 11)$

To remove the second set of parentheses, you must distribute the negative sign that is to its left. It changes every sign inside the parentheses.

$= 6y^2 + 2y + 5 - 5y^2 + 6y + 11$

Now you combine like terms.

$= 6y^2 - 5y^2 + 2y + 6y + 5 + 11$

$= y^2 + 8y + 16$

7 0

3 years ago

PLS HELP ASAP!!

zepelin [54]

Answer:

D

Step-by-step explanation:

$A = P(1 + \frac{r}{n})^{nt}\\$

[5000* (1+0.125)]/12 =234.375, rounds up to $234.38

8 0

3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

3 years ago